#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e17;
const int INF = 1e9;
const int N = 1234567;
const string V = " 2017";
struct Node {
  int a[5][5];
  Node() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++)
        if (i != j)
          a[i][j] = INF;
        else
          a[i][j] = 0;
    }
  }
  bool OK() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++)
        if (a[i][j] != (int)1e9) return 0;
    return 1;
  }
};
Node Merge(Node A, Node B) {
  Node C;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) C.a[i][j] = INF;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) {
      for (int k = 0; k <= j; k++)
        C.a[i][j] = min(C.a[i][j], A.a[i][k] + B.a[k][j]);
    }
  return C;
}
int n, q;
string s;
Node ans, it[N];
Node operator+(Node x, Node y) {
  Node c = Node();
  for (int i = 0; i < 5; i++) c.a[i][i] = INF;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      for (int l = 0; l < 5; l++) {
        c.a[i][j] = min(c.a[i][j], x.a[i][l] + y.a[l][j]);
      }
    }
  }
  return c;
}
void buildit(int k, int l, int r) {
  if (l == r) {
    it[k] = Node();
    if (s[l] == '2') {
      it[k].a[0][0] = 1;
      it[k].a[0][1] = 0;
    } else if (s[l] == '0') {
      it[k].a[1][1] = 1;
      it[k].a[1][2] = 0;
    } else if (s[l] == '1') {
      it[k].a[2][2] = 1;
      it[k].a[2][3] = 0;
    } else if (s[l] == '6') {
      it[k].a[3][3] = 1;
      it[k].a[4][4] = 1;
    } else if (s[l] == '7') {
      it[k].a[3][3] = 1;
      it[k].a[3][4] = 0;
    }
  } else {
    int mid = (l + r) / 2;
    buildit(2 * k, l, mid);
    buildit(2 * k + 1, mid + 1, r);
    it[k] = it[2 * k] + it[2 * k + 1];
  }
}
Node query(int k, int l, int r, int ll, int rr) {
  if (l > rr || ll > r) return Node();
  if (l >= ll && r <= rr) {
    return it[k];
  } else {
    int mid = (l + r) / 2;
    Node res =
        query(2 * k, l, mid, ll, rr) + query(2 * k + 1, mid + 1, r, ll, rr);
    return res;
  }
}
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q >> s;
  s = ' ' + s;
  buildit(1, 1, n);
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    Node ans = query(1, 1, n, l, r);
    if (ans.a[0][4] > (r - l + 1)) {
      cout << -1 << endl;
    } else {
      cout << ans.a[0][4] << endl;
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
const long long inf = 1e17;
const int INF = 1e9;
const int N = 1234567;
const string V = " 2017";
struct Node {
  int a[5][5];
  Node() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++)
        if (i != j)
          a[i][j] = INF;
        else
          a[i][j] = 0;
    }
  }
  bool OK() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++)
        if (a[i][j] != (int)1e9) return 0;
    return 1;
  }
};
Node Merge(Node A, Node B) {
  Node C;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) C.a[i][j] = INF;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) {
      for (int k = 0; k <= j; k++)
        C.a[i][j] = min(C.a[i][j], A.a[i][k] + B.a[k][j]);
    }
  return C;
}
int n, q;
string s;
Node ans, it[N];
Node operator+(Node x, Node y) {
  Node c = Node();
  for (int i = 0; i < 5; i++) c.a[i][i] = INF;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      for (int l = 0; l < 5; l++) {
        c.a[i][j] = min(c.a[i][j], x.a[i][l] + y.a[l][j]);
      }
    }
  }
  return c;
}
void buildit(int k, int l, int r) {
  if (l == r) {
    it[k] = Node();
    if (s[l] == '2') {
      it[k].a[0][0] = 1;
      it[k].a[0][1] = 0;
    } else if (s[l] == '0') {
      it[k].a[1][1] = 1;
      it[k].a[1][2] = 0;
    } else if (s[l] == '1') {
      it[k].a[2][2] = 1;
      it[k].a[2][3] = 0;
    } else if (s[l] == '6') {
      it[k].a[3][3] = 1;
      it[k].a[4][4] = 1;
    } else if (s[l] == '7') {
      it[k].a[3][3] = 1;
      it[k].a[3][4] = 0;
    }
  } else {
    int mid = (l + r) / 2;
    buildit(2 * k, l, mid);
    buildit(2 * k + 1, mid + 1, r);
    it[k] = it[2 * k] + it[2 * k + 1];
  }
}
Node query(int k, int l, int r, int ll, int rr) {
  if (l > rr || ll > r) return Node();
  if (l >= ll && r <= rr) {
    return it[k];
  } else {
    int mid = (l + r) / 2;
    Node res =
        query(2 * k, l, mid, ll, rr) + query(2 * k + 1, mid + 1, r, ll, rr);
    return res;
  }
}
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q >> s;
  s = ' ' + s;
  buildit(1, 1, n);
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    Node ans = query(1, 1, n, l, r);
    if (ans.a[0][4] > (r - l + 1)) {
      cout << -1 << endl;
    } else {
      cout << ans.a[0][4] << endl;
    }
  }
  return 0;
}

```