NTT的一个题目，T的范围是1e3，a和b的范围是1e5，a和b的绝对值之差不超过5，保证tot全程>0。

程序复杂度按理说是$O(T^2logT)$的，但4s都跑不过去就很离谱QAQ

#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
#define maxn 100005
using namespace std;
typedef long long ll;
const int mod = 998244353;
const int mx = 1e5;
int read() {
	int x = 0, f = 1, ch = getchar();
	while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
	while(isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
	return x * f;
}

int fac[maxn], inv[maxn];
int pw(int a, int b) {int res = 1; while(b) {if(b & 1) res = 1ll * res * a % mod; a = 1ll *  a * a % mod, b >>= 1;} return res;}
int r[maxn], len, l;
void NTT(int *c, int flag) {
	for(int i = 1; i <= len; i++) if(i < r[i]) swap(c[i], c[r[i]]);
	for(int mid = 1; mid < len; mid <<= 1) {
		int gn = pw(3, (mod - 1) / (mid << 1));
		if(flag == -1) gn = pw(gn, mod - 2);
		for(int ls = 0, L = mid << 1; ls < len; ls += L) {
			int g = 1;
			for(int k = 0; k < mid; k++, g = 1ll * g * gn % mod) {
				ll x = c[ls + k], y = 1ll * g * c[ls + mid + k] % mod;
				c[ls + k] = (1ll * x + y) % mod, c[ls + mid + k] = (1ll * x - y + mod) % mod;
			}
		}
	}
	int rev = pw(len, mod - 2);
	if(flag == -1) for(int i = 0; i <= len; i++) c[i] = 1ll * c[i] * rev % mod;
}

int T, a, b;
int g[maxn], f[maxn];
signed main() {
	fac[0] = inv[0] = 1;
	for(int i = 1; i <= mx; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
	inv[mx] = pw(fac[mx], mod - 2);
	for(int i = mx - 1; i; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
	
	register int tot = 0;
	T = read(); f[0] = 1;
	ll ans = 0;
	while(T--) {
		a = read(), b = read();
		len = 1, l = 0;
		while(len <= a - b + tot + tot + tot) len <<= 1, l++;
		for(int i = 0; i <= len; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << l - 1), g[i] = 0;
		
		for(int i = 0; i <= a - b + tot + tot; i++) if(b + (i - tot) >= 0 && a - (i - tot) >= 0)
			g[i] = 1ll * inv[b + i - tot] * inv[a - i + tot] % mod;
		for(int i = tot + 1; i <= len; i++) f[i] = 0;
		NTT(f, 1), NTT(g, 1);
		for(int i = 0; i <= len; i++) f[i] = 1ll * f[i] * g[i] % mod;
		NTT(f, -1);
		for(int i = 0; i <= tot + a - b; i++) f[i] = 1ll * f[i + tot] * fac[a + b] % mod;
		tot += a - b;
	}
	for(int i = 0; i <= tot; i++) ans = (ans + f[i]) % mod;
	printf("%lld\n", ans);
	return 0;
}