#include <bits/stdc++.h>
using namespace std;
int read() {
  int f = 1, x = 0;
  char c = getchar();
  while (c < '0' || '9' < c) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return f * x;
}
struct Matrix {
  int a[5][5];
  Matrix() { memset(a, 0x3f, sizeof(a)); }
  int* operator[](int x) { return a[x]; }
  friend Matrix operator*(Matrix A, Matrix B) {
    Matrix C;
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++)
        for (int k = 0; k < 5; k++) C[i][j] = min(C[i][j], A[i][k] + B[k][j]);
    return C;
  }
};
char s[200000 + 5];
Matrix t[4 * 200000 + 5];
void Build(int u, int L, int R) {
  if (L == R) {
    t[u][0][0] = t[u][1][1] = t[u][2][2] = t[u][3][3] = t[u][4][4] = 0;
    if (s[L] == '2')
      t[u][0][0] = 1, t[u][1][0] = 0;
    else if (s[L] == '0')
      t[u][1][1] = 1, t[u][2][1] = 0;
    else if (s[L] == '1')
      t[u][2][2] = 1, t[u][3][2] = 0;
    else if (s[L] == '7')
      t[u][3][3] = 1, t[u][4][3] = 0;
    else if (s[L] == '6')
      t[u][3][3] = 1, t[u][4][4] = 1;
    return;
  }
  int Mid = (L + R) >> 1;
  Build((u << 1), L, Mid), Build((u << 1 | 1), Mid + 1, R);
  t[u] = t[(u << 1 | 1)] * t[(u << 1)];
  return;
}
Matrix Query(int u, int L, int R, int qL, int qR) {
  if (qL <= L && R <= qR) return t[u];
  int Mid = (L + R) >> 1;
  if (qR <= Mid)
    return Query((u << 1), L, Mid, qL, qR);
  else if (Mid + 1 <= qL)
    return Query((u << 1 | 1), Mid + 1, R, qL, qR);
  return Query((u << 1 | 1), Mid + 1, R, qL, qR) *
         Query((u << 1), L, Mid, qL, qR);
}
int main() {
  int n = read(), q = read();
  scanf("%s", s + 1);
  Build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int l = read(), r = read();
    Matrix Ans = Query(1, 1, n, l, r);
    printf("%d\n", Ans[4][0] > (r - l + 1) ? -1 : Ans[4][0]);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

Input

8 3
20166766
1 8
1 7
2 8


Output

4
3
-1


Input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15


Output

-1
2
1
-1
-1


Input

4 2
1234
2 4
1 2


Output

-1
-1

Note

In the first sample:

  * In the first query, ugliness("20166766") = 4 because all four sixes must be removed. 
  * In the second query, ugliness("2016676") = 3 because all three sixes must be removed. 
  * In the third query, ugliness("0166766") = - 1 because it's impossible to remove some digits to get a nice string. 



In the second sample:

  * In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice. 
  * In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170". 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int read() {
  int f = 1, x = 0;
  char c = getchar();
  while (c < '0' || '9' < c) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return f * x;
}
struct Matrix {
  int a[5][5];
  Matrix() { memset(a, 0x3f, sizeof(a)); }
  int* operator[](int x) { return a[x]; }
  friend Matrix operator*(Matrix A, Matrix B) {
    Matrix C;
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++)
        for (int k = 0; k < 5; k++) C[i][j] = min(C[i][j], A[i][k] + B[k][j]);
    return C;
  }
};
char s[200000 + 5];
Matrix t[4 * 200000 + 5];
void Build(int u, int L, int R) {
  if (L == R) {
    t[u][0][0] = t[u][1][1] = t[u][2][2] = t[u][3][3] = t[u][4][4] = 0;
    if (s[L] == '2')
      t[u][0][0] = 1, t[u][1][0] = 0;
    else if (s[L] == '0')
      t[u][1][1] = 1, t[u][2][1] = 0;
    else if (s[L] == '1')
      t[u][2][2] = 1, t[u][3][2] = 0;
    else if (s[L] == '7')
      t[u][3][3] = 1, t[u][4][3] = 0;
    else if (s[L] == '6')
      t[u][3][3] = 1, t[u][4][4] = 1;
    return;
  }
  int Mid = (L + R) >> 1;
  Build((u << 1), L, Mid), Build((u << 1 | 1), Mid + 1, R);
  t[u] = t[(u << 1 | 1)] * t[(u << 1)];
  return;
}
Matrix Query(int u, int L, int R, int qL, int qR) {
  if (qL <= L && R <= qR) return t[u];
  int Mid = (L + R) >> 1;
  if (qR <= Mid)
    return Query((u << 1), L, Mid, qL, qR);
  else if (Mid + 1 <= qL)
    return Query((u << 1 | 1), Mid + 1, R, qL, qR);
  return Query((u << 1 | 1), Mid + 1, R, qL, qR) *
         Query((u << 1), L, Mid, qL, qR);
}
int main() {
  int n = read(), q = read();
  scanf("%s", s + 1);
  Build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int l = read(), r = read();
    Matrix Ans = Query(1, 1, n, l, r);
    printf("%d\n", Ans[4][0] > (r - l + 1) ? -1 : Ans[4][0]);
  }
  return 0;
}

```