#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 3;
const int OO = 1e9;
int n, q;
string str;
void print(int dp[][5]) {
  for (int(i) = (0); (i) < (5); (i)++) {
    for (int(j) = (0); (j) < (5); (j)++) {
      if (dp[i][j] >= OO)
        printf("OO  ");
      else
        printf("%-3d ", dp[i][j]);
    }
    printf("\n");
  }
}
struct Node {
  int dp[5][5];
  Node operator*(const Node& node) {
    Node ret;
    for (int(i) = (0); (i) < (5); (i)++) {
      for (int(j) = (0); (j) < (5); (j)++) {
        ret.dp[i][j] = OO;
        for (int(k) = (i); (k) < (j + 1); (k)++) {
          ret.dp[i][j] = min(ret.dp[i][j], dp[i][k] + node.dp[k][j]);
        }
      }
    }
    return ret;
  }
  void init(int pos) {
    for (int(i) = (0); (i) < (5); (i)++) {
      for (int(j) = (0); (j) < (5); (j)++) {
        dp[i][j] = OO;
      }
    }
    for (int(i) = (0); (i) < (5); (i)++) {
      dp[i][i] = 0;
      switch (str[pos]) {
        case '2':
          if (i == 0) {
            dp[i][i + 1] = 0;
            dp[i][i] = 1;
          }
          break;
        case '0':
          if (i == 1) {
            dp[i][i + 1] = 0;
            dp[i][i] = 1;
          }
          break;
        case '1':
          if (i == 2) {
            dp[i][i + 1] = 0;
            dp[i][i] = 1;
          }
          break;
        case '6':
          if (i == 3 || i == 4) {
            dp[i][i] = 1;
          }
          break;
        case '7':
          if (i == 3) {
            dp[i][i + 1] = 0;
            dp[i][i] = 1;
          }
          break;
      }
    }
  }
  static Node neutral() {
    Node ret;
    for (int(i) = (0); (i) < (5); (i)++) {
      for (int(j) = (0); (j) < (5); (j)++) {
        ret.dp[i][j] = OO;
        ret.dp[i][i] = 0;
      }
    }
    return ret;
  }
};
struct SegTree {
  vector<Node> t;
  int ds;
  SegTree(int n) {
    for (ds = 1; ds < n; ds *= 2)
      ;
    t = vector<Node>(2 * ds + 1);
  }
  virtual void build(const string& s, int v, int l, int r) {
    if (l == r) {
      if (l < n) {
        t[v].init(l);
      } else {
        t[v] = Node::neutral();
      }
    } else {
      int m = (l + r) / 2;
      build(s, v + v, l, m);
      build(s, v + v + 1, m + 1, r);
      t[v] = t[v + v] * t[v + v + 1];
    }
  }
  virtual Node qrec(int L, int R, int v, int l, int r) {
    if (L <= l && r <= R) {
      return t[v];
    } else if (r < L || l > R) {
      return Node::neutral();
    } else {
      int m = (l + r) / 2;
      return qrec(L, R, 2 * v, l, m) * qrec(L, R, 2 * v + 1, m + 1, r);
    }
  }
  Node query(int L, int R) { return qrec(L, R, 1, 0, ds - 1); }
  void build(const string& s) { build(s, 1, 0, ds - 1); }
};
int main(int argc, char* argv[]) {
  ios::sync_with_stdio(0);
  cin >> n >> q;
  cin >> str;
  SegTree t(n);
  t.build(str);
  for (int(i) = (0); (i) < (q); (i)++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    auto ans = t.query(a, b).dp[0][4];
    if (ans >= OO) ans = -1;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 = 1e6 + 3;
const int OO = 1e9;
int n, q;
string str;
void print(int dp[][5]) {
  for (int(i) = (0); (i) < (5); (i)++) {
    for (int(j) = (0); (j) < (5); (j)++) {
      if (dp[i][j] >= OO)
        printf("OO  ");
      else
        printf("%-3d ", dp[i][j]);
    }
    printf("\n");
  }
}
struct Node {
  int dp[5][5];
  Node operator*(const Node& node) {
    Node ret;
    for (int(i) = (0); (i) < (5); (i)++) {
      for (int(j) = (0); (j) < (5); (j)++) {
        ret.dp[i][j] = OO;
        for (int(k) = (i); (k) < (j + 1); (k)++) {
          ret.dp[i][j] = min(ret.dp[i][j], dp[i][k] + node.dp[k][j]);
        }
      }
    }
    return ret;
  }
  void init(int pos) {
    for (int(i) = (0); (i) < (5); (i)++) {
      for (int(j) = (0); (j) < (5); (j)++) {
        dp[i][j] = OO;
      }
    }
    for (int(i) = (0); (i) < (5); (i)++) {
      dp[i][i] = 0;
      switch (str[pos]) {
        case '2':
          if (i == 0) {
            dp[i][i + 1] = 0;
            dp[i][i] = 1;
          }
          break;
        case '0':
          if (i == 1) {
            dp[i][i + 1] = 0;
            dp[i][i] = 1;
          }
          break;
        case '1':
          if (i == 2) {
            dp[i][i + 1] = 0;
            dp[i][i] = 1;
          }
          break;
        case '6':
          if (i == 3 || i == 4) {
            dp[i][i] = 1;
          }
          break;
        case '7':
          if (i == 3) {
            dp[i][i + 1] = 0;
            dp[i][i] = 1;
          }
          break;
      }
    }
  }
  static Node neutral() {
    Node ret;
    for (int(i) = (0); (i) < (5); (i)++) {
      for (int(j) = (0); (j) < (5); (j)++) {
        ret.dp[i][j] = OO;
        ret.dp[i][i] = 0;
      }
    }
    return ret;
  }
};
struct SegTree {
  vector<Node> t;
  int ds;
  SegTree(int n) {
    for (ds = 1; ds < n; ds *= 2)
      ;
    t = vector<Node>(2 * ds + 1);
  }
  virtual void build(const string& s, int v, int l, int r) {
    if (l == r) {
      if (l < n) {
        t[v].init(l);
      } else {
        t[v] = Node::neutral();
      }
    } else {
      int m = (l + r) / 2;
      build(s, v + v, l, m);
      build(s, v + v + 1, m + 1, r);
      t[v] = t[v + v] * t[v + v + 1];
    }
  }
  virtual Node qrec(int L, int R, int v, int l, int r) {
    if (L <= l && r <= R) {
      return t[v];
    } else if (r < L || l > R) {
      return Node::neutral();
    } else {
      int m = (l + r) / 2;
      return qrec(L, R, 2 * v, l, m) * qrec(L, R, 2 * v + 1, m + 1, r);
    }
  }
  Node query(int L, int R) { return qrec(L, R, 1, 0, ds - 1); }
  void build(const string& s) { build(s, 1, 0, ds - 1); }
};
int main(int argc, char* argv[]) {
  ios::sync_with_stdio(0);
  cin >> n >> q;
  cin >> str;
  SegTree t(n);
  t.build(str);
  for (int(i) = (0); (i) < (q); (i)++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    auto ans = t.query(a, b).dp[0][4];
    if (ans >= OO) ans = -1;
    cout << ans << endl;
  }
  return 0;
}

```