There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143
#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const int MN = 3010;
const ll MOD = 1000000007;

inline void add(ll &x, ll y) { x += y; if (x >= MOD) x -= MOD; }

ll comb[MN][MN];
ll dp[MN][MN];
int tr[MN];
int num[MN];

int main() {
	rep(i, 3010) {
		comb[i][0] = comb[i][i] = 1;
		for (int j = 1; j < i; ++j) {
			comb[i][j] = (comb[i-1][j-1] + comb[i-1][j]) % MOD;
		}
	}

	int N, M;
	string s;

	scanf("%d %d", &N, &M);
	cin >> s;

	rep(i, N) tr[i] = i;
	rep(i, N) num[i+1] = num[i] + (s[i] == '0');

	rep(i, M) {
		int l, r;
		scanf("%d %d", &l, &r);
		--l; --r;

		for (int k = l; k <= r; ++k) {
			tr[k] = max(tr[k], r);
		}
	}

	dp[0][0] = 1;

	int la = -1;

	rep(i, N) {
		rep(j, MN) if (dp[i][j] != 0) {
			int z = j + num[tr[i]+1] - num[la+1];
			if (z) add(dp[i+1][z-1], dp[i][j]);
			if (z < tr[i] - i + 1) add(dp[i+1][z], dp[i][j]);
		}
		la = tr[i];
	}

	cout << dp[N][0] << endl;

	return 0;
}