#include <iostream>
#include <vector>
#include <stack>
#include <string>
#include <cctype>
#include <cassert>

using namespace std;

const int MOD = 1e9 + 7;

// Function to check if an expression is valid
bool isValidExpression(const string& expr) {
    stack<char> stk;
    char lastChar = '\0';
    for (char c : expr) {
        if (isdigit(c)) {
            if (lastChar != '\0' && isdigit(lastChar)) {
                return false;
            }
            lastChar = c;
        } else if (c == '+' || c == '-' || c == '*') {
            if (lastChar == '\0' || lastChar == '+' || lastChar == '-' || lastChar == '*' || lastChar == '(') {
                return false;
            }
            lastChar = c;
        } else if (c == '(') {
            stk.push(c);
            lastChar = c;
        } else if (c == ')') {
            if (lastChar == '+' || lastChar == '-' || lastChar == '*' || lastChar == '(') {
                return false;
            }
            if (stk.empty()) {
                return false;
            }
            stk.pop();
            lastChar = c;
        } else {
            return false;
        }
    }
    return lastChar != '\0' && !isdigit(lastChar) && stk.empty();
}

// Function to evaluate the expression
long long evaluateExpression(const string& expr) {
    stack<long long> values;
    stack<char> ops;

    auto precedence = [](char op) -> int {
        if (op == '+' || op == '-') return 1;
        if (op == '*') return 2;
        return 0;
    };

    auto applyOp = [](long long a, long long b, char op) -> long long {
        if (op == '+') return (a + b) % MOD;
        if (op == '-') return (a - b + MOD) % MOD;
        if (op == '*') return (a * b) % MOD;
        return 0;
    };

    for (char c : expr) {
        if (isdigit(c)) {
            values.push(c - '0');
        } else if (c == '(') {
            ops.push(c);
        } else if (c == ')') {
            while (!ops.empty() && ops.top() != '(') {
                long long val2 = values.top(); values.pop();
                long long val1 = values.top(); values.pop();
                char op = ops.top(); ops.pop();
                values.push(applyOp(val1, val2, op));
            }
            ops.pop(); // remove '('
        } else if (c == '+' || c == '-' || c == '*') {
            while (!ops.empty() && precedence(ops.top()) >= precedence(c)) {
                long long val2 = values.top(); values.pop();
                long long val1 = values.top(); values.pop();
                char op = ops.top(); ops.pop();
                values.push(applyOp(val1, val2, op));
            }
            ops.push(c);
        }
    }

    while (!ops.empty()) {
        long long val2 = values.top(); values.pop();
        long long val1 = values.top(); values.pop();
        char op = ops.top(); ops.pop();
        values.push(applyOp(val1, val2, op));
    }

    return values.top();
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m;
    cin >> n >> m;
    string expr;
    cin >> expr;

    vector<string> operations(m);
    for (int i = 0; i < m; ++i) {
        int type;
        cin >> type;
        if (type == 1) {
            int x;
            char y;
            cin >> x >> y;
            --x; // Convert to 0-based index
            expr[x] = y;
        } else if (type == 2) {
            int l, r;
            cin >> l >> r;
            --l; --r; // Convert to 0-based index
            string subExpr = expr.substr(l, r - l + 1);
            if (isValidExpression(subExpr)) {
                cout << evaluateExpression(subExpr) << '\n';
            } else {
                cout << -1 << '\n';
            }
        }
    }

    return 0;
}

P5067 [Ynoi2014] 长存不灭的过去、逐渐消逝的未来该如何解决？