#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 3005, mod = 1e9+7;

ll n, m, o[N], b[N], dt[N][N];
char a[N];

int main()
{
	scanf("%lld%lld%s",&n,&m,a+1);
	for(ll i=1;i<=n;i++) {
		o[i] = o[i-1] + (a[i] == '1');
	}
	iota(b, b+n+1, 0);
	for(ll i=1;i<=m;i++) {
		ll S, E;
		scanf("%lld%lld",&S,&E);
		for(ll i=S;i<=E;i++) b[i] = max(b[i], E);
	}
	dt[0][0] = 1;
	for(ll i=1;i<=n;i++) for(ll j=0;j<=i;j++) {
		if(j > o[b[i]] || i-j > b[i]-o[b[i]]) continue;
		dt[i][j] = dt[i-1][j];
		if(j) dt[i][j] = (dt[i][j] + dt[i-1][j-1]) % mod;
	}
	printf("%lld\n",dt[n][o[n]]);
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
typedef long long ll;
const ll N = 3005, mod = 1e9+7;

ll n, m, o[N], b[N], dt[N][N];
char a[N];

int main()
{
	scanf("%lld%lld%s",&n,&m,a+1);
	for(ll i=1;i<=n;i++) {
		o[i] = o[i-1] + (a[i] == '1');
	}
	iota(b, b+n+1, 0);
	for(ll i=1;i<=m;i++) {
		ll S, E;
		scanf("%lld%lld",&S,&E);
		for(ll i=S;i<=E;i++) b[i] = max(b[i], E);
	}
	dt[0][0] = 1;
	for(ll i=1;i<=n;i++) for(ll j=0;j<=i;j++) {
		if(j > o[b[i]] || i-j > b[i]-o[b[i]]) continue;
		dt[i][j] = dt[i-1][j];
		if(j) dt[i][j] = (dt[i][j] + dt[i-1][j-1]) % mod;
	}
	printf("%lld\n",dt[n][o[n]]);
}

```