#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, M = 5;
int a[N], n, Q;
struct Node {
  int mat[M][M];
  Node() {}
  Node(int x) {
    memset(mat, 0x3f, sizeof mat);
    for (int i = 0; i <= M - 1; ++i) mat[i][i] = 0;
    if (x == 2)
      mat[0][0] = 1, mat[0][1] = 0;
    else if (x == 0)
      mat[1][1] = 1, mat[1][2] = 0;
    else if (x == 1)
      mat[2][2] = 1, mat[2][3] = 0;
    else if (x == 7)
      mat[3][3] = 1, mat[3][4] = 0;
    else if (x == 6)
      mat[3][3] = 1, mat[4][4] = 1;
  }
  friend Node operator*(Node a, Node b) {
    Node c(-1);
    memset(c.mat, 0x3f, sizeof mat);
    for (int i = 0; i <= M - 1; ++i)
      for (int j = 0; j <= M - 1; ++j)
        for (int k = 0; k <= M - 1; ++k)
          c.mat[i][j] = min(c.mat[i][j], a.mat[i][k] + b.mat[k][j]);
    return c;
  }
} t[N << 2];
inline void build(int k, int l, int r) {
  if (l == r) {
    t[k] = Node(a[l]);
    return;
  }
  int md = l + r >> 1;
  build(k << 1, l, md), build(k << 1 | 1, md + 1, r);
  t[k] = t[k << 1] * t[k << 1 | 1];
}
Node query(int k, int l, int r, int x, int y) {
  if (x <= l && r <= y) return t[k];
  int md = l + r >> 1;
  Node res(-1);
  if (x <= md) res = res * query(k << 1, l, md, x, y);
  if (y > md) res = res * query(k << 1 | 1, md + 1, r, x, y);
  return res;
}
inline int query(int x, int y) {
  int an = query(1, 1, n, x, y).mat[0][4];
  return an > n ? -1 : an;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> Q;
  for (int i = 1; i <= n; ++i) {
    char x;
    cin >> x;
    a[i] = x - '0';
  }
  build(1, 1, n);
  for (int i = 1; i <= Q; ++i) {
    int x, y;
    cin >> x >> y;
    cout << query(x, y) << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 = 2e5 + 5, M = 5;
int a[N], n, Q;
struct Node {
  int mat[M][M];
  Node() {}
  Node(int x) {
    memset(mat, 0x3f, sizeof mat);
    for (int i = 0; i <= M - 1; ++i) mat[i][i] = 0;
    if (x == 2)
      mat[0][0] = 1, mat[0][1] = 0;
    else if (x == 0)
      mat[1][1] = 1, mat[1][2] = 0;
    else if (x == 1)
      mat[2][2] = 1, mat[2][3] = 0;
    else if (x == 7)
      mat[3][3] = 1, mat[3][4] = 0;
    else if (x == 6)
      mat[3][3] = 1, mat[4][4] = 1;
  }
  friend Node operator*(Node a, Node b) {
    Node c(-1);
    memset(c.mat, 0x3f, sizeof mat);
    for (int i = 0; i <= M - 1; ++i)
      for (int j = 0; j <= M - 1; ++j)
        for (int k = 0; k <= M - 1; ++k)
          c.mat[i][j] = min(c.mat[i][j], a.mat[i][k] + b.mat[k][j]);
    return c;
  }
} t[N << 2];
inline void build(int k, int l, int r) {
  if (l == r) {
    t[k] = Node(a[l]);
    return;
  }
  int md = l + r >> 1;
  build(k << 1, l, md), build(k << 1 | 1, md + 1, r);
  t[k] = t[k << 1] * t[k << 1 | 1];
}
Node query(int k, int l, int r, int x, int y) {
  if (x <= l && r <= y) return t[k];
  int md = l + r >> 1;
  Node res(-1);
  if (x <= md) res = res * query(k << 1, l, md, x, y);
  if (y > md) res = res * query(k << 1 | 1, md + 1, r, x, y);
  return res;
}
inline int query(int x, int y) {
  int an = query(1, 1, n, x, y).mat[0][4];
  return an > n ? -1 : an;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> Q;
  for (int i = 1; i <= n; ++i) {
    char x;
    cin >> x;
    a[i] = x - '0';
  }
  build(1, 1, n);
  for (int i = 1; i <= Q; ++i) {
    int x, y;
    cin >> x >> y;
    cout << query(x, y) << '\n';
  }
  return 0;
}

```