#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
const int INF = 0x3f3f3f3f;
struct mat {
  int a[5][5];
  int* operator[](int x) { return a[x]; }
  void init(int x) {
    for (int i = (0); i <= (4); ++i)
      for (int j = (0); j <= (4); ++j) a[i][j] = i == j ? 0 : INF;
    switch (x) {
      case 2:
        a[0][0] = 1;
        a[0][1] = 0;
        break;
      case 0:
        a[1][1] = 1;
        a[1][2] = 0;
        break;
      case 1:
        a[2][2] = 1;
        a[2][3] = 0;
        break;
      case 6:
        a[3][3] = 1;
        a[4][4] = 1;
        break;
      case 7:
        a[3][3] = 1;
        a[3][4] = 0;
        break;
    }
  }
  mat operator+(mat A) const {
    mat B;
    for (int i = (0); i <= (4); ++i)
      for (int j = (0); j <= (4); ++j) B.a[i][j] = INF;
    for (int i = (0); i <= (4); ++i)
      for (int j = (i); j <= (4); ++j)
        for (int k = (i); k <= (j); ++k)
          B.a[i][j] = min(B.a[i][j], a[i][k] + A.a[k][j]);
    return B;
  }
} T[N << 2];
int n, l, r, q;
char s[N];
inline void build(int p, int l, int r) {
  if (l == r) {
    T[p].init(s[l]);
    return;
  }
  int mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  T[p] = T[p << 1] + T[p << 1 | 1];
}
mat ask(int p, int l, int r, int L, int R) {
  if (l == L && r == R) return T[p];
  int mid = (l + r) >> 1;
  if (R <= mid)
    return ask(p << 1, l, mid, L, R);
  else if (L > mid)
    return ask(p << 1 | 1, mid + 1, r, L, R);
  return ask(p << 1, l, mid, L, mid) + ask(p << 1 | 1, mid + 1, r, mid + 1, R);
}
int main() {
  scanf("%d%d", &n, &q);
  scanf(" %s", s + 1);
  for (int i = (1); i <= (n); ++i) s[i] -= '0';
  build(1, 1, n);
  while (q--) {
    scanf("%d%d", &l, &r);
    mat A = ask(1, 1, n, l, r);
    printf("%d\n", A[0][4] >= INF ? -1 : A[0][4]);
  }
}

### 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;
const int N = 200010;
const int INF = 0x3f3f3f3f;
struct mat {
  int a[5][5];
  int* operator[](int x) { return a[x]; }
  void init(int x) {
    for (int i = (0); i <= (4); ++i)
      for (int j = (0); j <= (4); ++j) a[i][j] = i == j ? 0 : INF;
    switch (x) {
      case 2:
        a[0][0] = 1;
        a[0][1] = 0;
        break;
      case 0:
        a[1][1] = 1;
        a[1][2] = 0;
        break;
      case 1:
        a[2][2] = 1;
        a[2][3] = 0;
        break;
      case 6:
        a[3][3] = 1;
        a[4][4] = 1;
        break;
      case 7:
        a[3][3] = 1;
        a[3][4] = 0;
        break;
    }
  }
  mat operator+(mat A) const {
    mat B;
    for (int i = (0); i <= (4); ++i)
      for (int j = (0); j <= (4); ++j) B.a[i][j] = INF;
    for (int i = (0); i <= (4); ++i)
      for (int j = (i); j <= (4); ++j)
        for (int k = (i); k <= (j); ++k)
          B.a[i][j] = min(B.a[i][j], a[i][k] + A.a[k][j]);
    return B;
  }
} T[N << 2];
int n, l, r, q;
char s[N];
inline void build(int p, int l, int r) {
  if (l == r) {
    T[p].init(s[l]);
    return;
  }
  int mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  T[p] = T[p << 1] + T[p << 1 | 1];
}
mat ask(int p, int l, int r, int L, int R) {
  if (l == L && r == R) return T[p];
  int mid = (l + r) >> 1;
  if (R <= mid)
    return ask(p << 1, l, mid, L, R);
  else if (L > mid)
    return ask(p << 1 | 1, mid + 1, r, L, R);
  return ask(p << 1, l, mid, L, mid) + ask(p << 1 | 1, mid + 1, r, mid + 1, R);
}
int main() {
  scanf("%d%d", &n, &q);
  scanf(" %s", s + 1);
  for (int i = (1); i <= (n); ++i) s[i] -= '0';
  build(1, 1, n);
  while (q--) {
    scanf("%d%d", &l, &r);
    mat A = ask(1, 1, n, l, r);
    printf("%d\n", A[0][4] >= INF ? -1 : A[0][4]);
  }
}

```