#include <bits/stdc++.h>
using namespace std;
int n, q, inf = 1e9;
int liczby[(1 << 18) + 2];
struct node {
  int start, stop;
  int DB[5][5];
  node() {
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) DB[i][j] = inf;
  }
};
node ALL;
node mergex(node &left, node &right) {
  node res;
  for (int x = 0; x < 5; ++x)
    for (int z = x; z < 5; ++z)
      for (int y = z; y < 5; ++y) {
        res.DB[x][y] = min(res.DB[x][y], left.DB[x][z] + right.DB[z][y]);
      }
  return res;
}
struct segment_tree {
  node *T;
  segment_tree() { T = new node[2 * (1 << 18)]; }
  ~segment_tree() { delete[] T; }
  node build(int i, int start, int stop) {
    T[i].start = start;
    T[i].stop = stop;
    if (start == stop) {
      node &temp = T[i];
      for (int x = 0; x < 5; ++x) temp.DB[x][x] = 0;
      if (liczby[start] == 2) {
        temp.DB[0][0] = 1;
        temp.DB[0][1] = 0;
      } else if (liczby[start] == 0) {
        temp.DB[1][1] = 1;
        temp.DB[1][2] = 0;
      } else if (liczby[start] == 1) {
        temp.DB[2][2] = 1;
        temp.DB[2][3] = 0;
      } else if (liczby[start] == 7) {
        temp.DB[3][3] = 1;
        temp.DB[3][4] = 0;
      } else if (liczby[start] == 6) {
        temp.DB[3][3] = 1;
        temp.DB[4][4] = 1;
      }
      return temp;
    }
    int s = (start + stop) / 2;
    node left = build(i * 2, start, s);
    node right = build(i * 2 + 1, s + 1, stop);
    node temp = mergex(left, right);
    for (int x = 0; x < 5; ++x)
      for (int y = 0; y < 5; ++y) T[i].DB[x][y] = temp.DB[x][y];
    return T[i];
  }
  node query(int i, int a, int b) {
    int start = T[i].start;
    int stop = T[i].stop;
    if (a <= start and b >= stop) {
      return T[i];
    }
    if (a > stop or b < start) {
      return ALL;
    }
    node left = query(i * 2, a, b);
    node right = query(i * 2 + 1, a, b);
    return mergex(left, right);
  }
};
segment_tree tree;
int policz(int a, int b) {
  node temp = tree.query(1, a, b);
  if (temp.DB[0][4] >= inf) return -1;
  return temp.DB[0][4];
}
void init() {
  for (int i = 0; i < (1 << 18) + 1; ++i) liczby[i] = 8;
  for (int i = 0; i < 5; ++i) ALL.DB[i][i] = 0;
}
int main() {
  scanf("%d%d\n", &n, &q);
  init();
  for (int i = 1; i <= n; ++i) {
    char temp;
    temp = getchar();
    liczby[i] = temp - '0';
  }
  tree.build(1, 1, (1 << 18));
  for (int i = 0; i < q; ++i) {
    int a, b;
    cin >> a >> b;
    printf("%d\n", policz(a, b));
  }
}

### Prompt

Please create a solution in CPP to 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 n, q, inf = 1e9;
int liczby[(1 << 18) + 2];
struct node {
  int start, stop;
  int DB[5][5];
  node() {
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) DB[i][j] = inf;
  }
};
node ALL;
node mergex(node &left, node &right) {
  node res;
  for (int x = 0; x < 5; ++x)
    for (int z = x; z < 5; ++z)
      for (int y = z; y < 5; ++y) {
        res.DB[x][y] = min(res.DB[x][y], left.DB[x][z] + right.DB[z][y]);
      }
  return res;
}
struct segment_tree {
  node *T;
  segment_tree() { T = new node[2 * (1 << 18)]; }
  ~segment_tree() { delete[] T; }
  node build(int i, int start, int stop) {
    T[i].start = start;
    T[i].stop = stop;
    if (start == stop) {
      node &temp = T[i];
      for (int x = 0; x < 5; ++x) temp.DB[x][x] = 0;
      if (liczby[start] == 2) {
        temp.DB[0][0] = 1;
        temp.DB[0][1] = 0;
      } else if (liczby[start] == 0) {
        temp.DB[1][1] = 1;
        temp.DB[1][2] = 0;
      } else if (liczby[start] == 1) {
        temp.DB[2][2] = 1;
        temp.DB[2][3] = 0;
      } else if (liczby[start] == 7) {
        temp.DB[3][3] = 1;
        temp.DB[3][4] = 0;
      } else if (liczby[start] == 6) {
        temp.DB[3][3] = 1;
        temp.DB[4][4] = 1;
      }
      return temp;
    }
    int s = (start + stop) / 2;
    node left = build(i * 2, start, s);
    node right = build(i * 2 + 1, s + 1, stop);
    node temp = mergex(left, right);
    for (int x = 0; x < 5; ++x)
      for (int y = 0; y < 5; ++y) T[i].DB[x][y] = temp.DB[x][y];
    return T[i];
  }
  node query(int i, int a, int b) {
    int start = T[i].start;
    int stop = T[i].stop;
    if (a <= start and b >= stop) {
      return T[i];
    }
    if (a > stop or b < start) {
      return ALL;
    }
    node left = query(i * 2, a, b);
    node right = query(i * 2 + 1, a, b);
    return mergex(left, right);
  }
};
segment_tree tree;
int policz(int a, int b) {
  node temp = tree.query(1, a, b);
  if (temp.DB[0][4] >= inf) return -1;
  return temp.DB[0][4];
}
void init() {
  for (int i = 0; i < (1 << 18) + 1; ++i) liczby[i] = 8;
  for (int i = 0; i < 5; ++i) ALL.DB[i][i] = 0;
}
int main() {
  scanf("%d%d\n", &n, &q);
  init();
  for (int i = 1; i <= n; ++i) {
    char temp;
    temp = getchar();
    liczby[i] = temp - '0';
  }
  tree.build(1, 1, (1 << 18));
  for (int i = 0; i < q; ++i) {
    int a, b;
    cin >> a >> b;
    printf("%d\n", policz(a, b));
  }
}

```