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

#define FOR(i,l,r) for(int i = (l);i < (r);i++)
#define ALL(x) (x).begin(),(x).end()
template<typename T> bool chmax(T& a,const T& b){return a < b ? (a = b,true) : false;}
template<typename T> bool chmin(T& a,const T& b){return b < a ? (a = b,true) : false;}
typedef long long ll;

int N,M;
string S;
ll dp [3001] [3001];
const ll MOD = 1e9 + 7;

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

	cin >> N >> M >> S;
	vector< pair<int,int> > A(M);
	FOR(i,0,M){
		cin >> A [i].first >> A [i].second;
		A [i].first--;
	}

	dp [0] [0] = 1;
	int idx = 0,r = 0;
	FOR(i,0,N){
		int cnt = 0;
		if(r == i){
			cnt += S [i] == '0';
			r++;
		}
		while(A [idx].first <= i){
			FOR(j,r,A [idx].second){
				cnt += S [j] == '0';
			}
			chmax(r,A [idx].second);
			idx++;
		}
		for(int j = 0;j <= N;j++) if(dp [i] [j]){
			if(j + cnt > 0){
				(dp [i + 1] [j + cnt - 1] += dp [i] [j]) %= MOD;
			}
			if((r - i) - (j + cnt) > 0){
				(dp [i + 1] [j + cnt] += dp [i] [j]) %= MOD;
			}
		}
	}

	printf("%lld\n",dp [N] [0]);


	return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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

### Response

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

#define FOR(i,l,r) for(int i = (l);i < (r);i++)
#define ALL(x) (x).begin(),(x).end()
template<typename T> bool chmax(T& a,const T& b){return a < b ? (a = b,true) : false;}
template<typename T> bool chmin(T& a,const T& b){return b < a ? (a = b,true) : false;}
typedef long long ll;

int N,M;
string S;
ll dp [3001] [3001];
const ll MOD = 1e9 + 7;

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

	cin >> N >> M >> S;
	vector< pair<int,int> > A(M);
	FOR(i,0,M){
		cin >> A [i].first >> A [i].second;
		A [i].first--;
	}

	dp [0] [0] = 1;
	int idx = 0,r = 0;
	FOR(i,0,N){
		int cnt = 0;
		if(r == i){
			cnt += S [i] == '0';
			r++;
		}
		while(A [idx].first <= i){
			FOR(j,r,A [idx].second){
				cnt += S [j] == '0';
			}
			chmax(r,A [idx].second);
			idx++;
		}
		for(int j = 0;j <= N;j++) if(dp [i] [j]){
			if(j + cnt > 0){
				(dp [i + 1] [j + cnt - 1] += dp [i] [j]) %= MOD;
			}
			if((r - i) - (j + cnt) > 0){
				(dp [i + 1] [j + cnt] += dp [i] [j]) %= MOD;
			}
		}
	}

	printf("%lld\n",dp [N] [0]);


	return 0;
}

```