Problem: Consider a road consisting of several rows. Each row is divided into several rectangular tiles, and all tiles in the same row are equal. The first row contains exactly one rectangular tile. Look at the picture below which shows how the tiles are arranged.

The road is constructed as follows: 

  * the first row consists of 1 tile; 
  * then a_1 rows follow; each of these rows contains 1 tile greater than the previous row; 
  * then b_1 rows follow; each of these rows contains 1 tile less than the previous row; 
  * then a_2 rows follow; each of these rows contains 1 tile greater than the previous row; 
  * then b_2 rows follow; each of these rows contains 1 tile less than the previous row; 
  * ... 
  * then a_n rows follow; each of these rows contains 1 tile greater than the previous row; 
  * then b_n rows follow; each of these rows contains 1 tile less than the previous row. 

<image> An example of the road with n = 2, a_1 = 4, b_1 = 2, a_2 = 2, b_2 = 3. Rows are arranged from left to right. 

You start from the only tile in the first row and want to reach the last row (any tile of it). From your current tile, you can move to any tile in the next row which touches your current tile.

Calculate the number of different paths from the first row to the last row. Since it can be large, print it modulo 998244353.

Input

The first line contains one integer n (1 ≤ n ≤ 1000).

Then n lines follow. The i-th of them contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 10^5; |a_i - b_i| ≤ 5).

Additional constraint on the input: the sequence of a_i and b_i never results in a row with non-positive number of tiles.

Output

Print one integer — the number of paths from the first row to the last row, taken modulo 998244353.

Examples

Input


2
4 2
2 3


Output


850


Input


3
4 1
2 3
3 1


Output


10150


Input


8
328 323
867 868
715 718
721 722
439 435
868 870
834 834
797 796


Output


759099319

Solution (in CPP):

Solution:
/*
   ___                         ______      __                  __
   /   |____  __  ___________ _/ ____/___ _/ /___  ____  ______/ /____
   / /| /_  / / / / / ___/ __ `/ /   / __ `/ __/ / / / / / / __  / ___/
   / ___ |/ /_/ /_/ (__  ) /_/ / /___/ /_/ / /_/ /_/ / /_/ / /_/ (__  )
   /_/  |_/___/\__,_/____/\__,_/\____/\__,_/\__/\__, /\__, /\__,_/____/
   /____//____/

   　　　　　／＞　 フ
   　　　　　| 　_　 _|
   　 　　　／`ミ _x 彡
   　　 　 /　　　 　 |
   　　　 /　  ヽ　　 ?
   　／￣|　　 |　|　|
   　| (￣ヽ＿_ヽ_)_)
   　＼二つ

*/

#include <queue>
#include <vector>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MP make_pair
#define ll long long
#define fi first
#define se second
using namespace std;

template <typename T>
void read(T &x) {
	x = 0; bool f = 0;
	char c = getchar();
	for (;!isdigit(c);c=getchar()) if (c=='-') f=1;
	for (;isdigit(c);c=getchar()) x=x*10+(c^48);
	if (f) x=-x;
}

template<typename F>
inline void write(F x, char ed = '\n') {
	static short st[30];short tp=0;
	if(x<0) putchar('-'),x=-x;
	do st[++tp]=x%10,x/=10; while(x);
	while(tp) putchar('0'|st[tp--]);
	putchar(ed);
}

template <typename T>
inline void Mx(T &x, T y) { x < y && (x = y); }

template <typename T>
inline void Mn(T &x, T y) { x > y && (x = y); }

//#pragma GCC optimize(2)
const int P = 998244353;
namespace Poly {
#define poly vector<int> 
#define reg register
	inline int mod(int x) { return x >= P ? x - P : x; }
	inline void Add(int &x, int y) { x += y; if (x >= P) x -= P; }
	template <typename T>
		inline int max(T x, T y) { return x > y ? x : y; }
	template <typename T>
		inline int min(T x, T y) { return x > y ? y : x; }
	int fpw(ll x, int mi) {
		ll res = 1;
		for (; mi; mi >>= 1, x = x * x % P)
			if (mi & 1) res = res * x % P;
		return res;
	}
	const int N = 600500;
	int E[N], r[N], lim;
	inline void Prework(int Maxsize) {
		E[1] = lim = 1;
		while (lim <= Maxsize) lim <<= 1;
		for (int i = 2;i < lim;i <<= 1) {
			int *e0 = E + (i >> 1), *e1 = E + i;
			ll w = fpw(3, (P - 1) / (i << 1));
			for (int j = 0;j < i; j += 2)
				e1[j] = e0[j >> 1], e1[j + 1] = e1[j] * w % P;
		}
	}

	inline void getR(int len = lim >> 1) {
		for (reg int i = 1;i < lim; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) ? len : 0);
	}
	inline int getlen(int Maxsize) { lim = 1; while (lim <= Maxsize) lim <<= 1; return lim; } 
	void init(int Maxsize) { getlen(Maxsize); getR(); }

	inline void Dft(poly &A) {
		for (reg int i = 1;i < lim; ++i)
			if (r[i] < i) std::swap(A[i], A[r[i]]);
		if (lim >= 2) 
			for (reg int j = 0;j < lim; j += 2) {
				int x = A[j], y = A[j+1];
				A[j] = mod(x + y), A[j+1] = mod(P + x - y);
			}
		for (int i = 2;i < lim; i <<= 1) {
			for (int j = 0;j < lim; j += (i << 1)) {
				poly::iterator f = A.begin() + j, g = A.begin() + i + j;
				int *e = E + i;
				for (int k = 0;k < i; ++k) {
					int x = f[k], y = 1ll * g[k] * e[k] % P;
					f[k] = mod(x + y), g[k] = mod(P + x - y);
					++k;
					x = f[k], y = 1ll * g[k] * e[k] % P;
					f[k] = mod(x + y), g[k] = mod(P + x - y);
				}
			}
		}
	}

	inline void iDft(poly &f) { 
		Dft(f), reverse(f.begin() + 1, f.begin() + lim); 
		ll inv = P - (P - 1) / lim;
		for (int i = 0;i < lim; i++) f[i] = f[i] * inv % P;
	}

	poly Mul(poly &a, poly &b) {
		int s1 = a.size(), s2 = b.size();
		if (s1 <= 28 || s2 <= 28) {
			poly ans(s1 + s2 - 1);
			for (reg int i = 0;i < s1 + s2 - 1; ++i) ans[i] = 0;
			for (reg int i = 0;i < s1; ++i)
				for (reg int j = 0;j < s2; ++j)
					Add(ans[i + j], 1ll * a[i] * b[j] % P);
			return ans;
		}
		init(s1 + s2 - 2);
		poly ans(lim); a.resize(lim), b.resize(lim);
		Dft(a), Dft(b);
		for (int i = 0;i < lim; ++i) ans[i] = 1ll * a[i] * b[i] % P;
		return iDft(ans), ans.resize(s1 + s2 - 1), ans;
	}
	poly operator * (poly a, poly b) { return Mul(a, b); }
}

using Poly::Prework;
using Poly::operator*;

const int N = 205000;
int n;
ll fac[N], inv[N];
int main() {
	read(n), fac[0] = fac[1] = inv[0] = inv[1] = 1;
	for (int i = 2;i <= 200000; ++i) 
		fac[i] = fac[i-1] * i % P, inv[i] = (P - P / i) * inv[P % i] % P;
	for (int i = 1;i <= 200000; ++i)
		inv[i] = inv[i-1] * inv[i] % P;
	auto C = [&](int x, int y) {
		if (y > x || y < 0) return 0ll;
		return fac[x] * inv[y] % P * inv[x-y] % P;
	};
	Prework(10000);
	poly st(2); st[1] = 1;
	int nw = 1, tnw;
	for (int i = 1, a, b;i <= n; ++i) {
		read(a), read(b), tnw = nw + a - b;
		poly res(tnw + nw + 1);
		for (int i = 0;i <= tnw + nw; ++i)  res[i] = C(a + b, a + nw - i);
		poly T = st * res;
		st.clear();
		st.resize(tnw + 1);
		for (int i = 1;i <= tnw; ++i) st[i] = T[i + nw];
		nw = tnw; 
	}
	ll ans = 0;
	for (int i = 1;i <= nw; ++i) ans += st[i];
	write(ans % P);
	return 0;
}

/*

3
24 21
66 64
1 1

*/