Problem: JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.

First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.

For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.

However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.

All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).

The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.

Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.

Output

Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.

Examples

Input

4 2
1011
1 4
3 4


Output

14
3


Input

3 2
111
1 2
3 3


Output

3
1

Note

In the first example: 

  * For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2. 
  * For query 2: Both 3, 4 and 4, 3 ordering give the same answer. 



In the second example, any order of eating parts leads to the same answer.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  long long int r, i;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long int power(long long int x, long long int y, long long int mod) {
  long long int temp, ty, my;
  if (y == 0) return 1;
  temp = power(x, y / 2, mod);
  ty = (temp % mod) * (temp % mod);
  if (y % 2 == 0) {
    return ty % mod;
  } else {
    my = (x % mod) * (ty % mod);
    return my % mod;
  }
}
long long int mycbrt(long long int n) {
  long long int start = 0;
  long long int end = 1000000LL;
  long long int ans = -1;
  while (start <= end) {
    long long int mid = (start + end) / 2;
    if ((mid * mid * mid) > n)
      end = mid - 1;
    else if ((mid * mid * mid) == n) {
      ans = mid;
      break;
    } else {
      start = mid + 1;
    }
  }
  if (n == 1) {
    ans = 1;
  }
  return ans;
}
void SieveOfEratosthenes(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * 2; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) cout << p << " ";
}
struct abhi {
  long long int val1;
  long long int val2;
  long long int po;
};
bool cmp(struct abhi x, struct abhi y) {
  if (x.val1 == y.val1) return x.val2 < y.val2;
  return x.val1 < y.val1;
}
void fastscan(int &number) {
  bool negative = false;
  register int c;
  number = 0;
  c = getchar();
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
  if (negative) number *= -1;
}
vector<pair<long long int, long long int> > po;
long long int dp[2000010];
long long int dp2[200010];
long long int ar[2000010];
vector<long long int> vec;
vector<pair<long long int, long long int> > vec2;
long long int mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, i, j, k, q;
  long long int t;
  cin >> n >> t;
  string str;
  cin >> str;
  dp[0] = 0;
  for (i = (0); i < (n); i++) {
    dp[i + 1] = dp[i] + ((str[i] == '1') ? 1 : 0);
  }
  while (t--) {
    long long int a, b;
    cin >> a >> b;
    long long int ones = dp[b] - dp[a - 1];
    long long int zeroes = (b - a + 1) - ones;
    long long int pot = (power(2, ones, mod) - 1 + mod) % mod;
    long long int ans = pot;
    long long int pot2 = (power(2, zeroes, mod) - 1 + mod) % mod;
    long long int ko = (pot * pot2) % mod;
    ans = (ans + ko) % mod;
    cout << ans << "\n";
  }
  return 0;
}