#include <bits/stdc++.h>
using namespace std;
const int Inf = 1e9 + 7;
const int N = 2e5 + 7;
int n;
string s;
struct Node {
  int m[5][5];
  int *operator[](int x) { return m[x]; }
  Node() : Node('?') {}
  Node(char x) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) m[i][j] = (i != j ? Inf : 0);
    switch (x) {
      case '2':
        m[0][0] = 1;
        m[0][1] = 0;
        break;
      case '0':
        m[1][1] = 1;
        m[1][2] = 0;
        break;
      case '1':
        m[2][2] = 1;
        m[2][3] = 0;
        break;
      case '7':
        m[3][3] = 1;
        m[3][4] = 0;
        break;
      case '6':
        m[3][3] = 1;
        m[4][4] = 1;
      default:
        break;
    }
  }
};
Node tree[4 * N];
Node f(const Node &a, const Node &b) {
  Node res;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      res[i][j] = Inf;
      for (int k = 0; k < 5; k++)
        res[i][j] = min(res[i][j], a.m[i][k] + b.m[k][j]);
    }
  }
  return res;
}
void build(int v = 1, int l = 0, int r = n) {
  if (l + 1 == r) {
    tree[v] = Node(s[l]);
    return;
  }
  int md = (l + r) / 2;
  build(2 * v, l, md);
  build(2 * v + 1, md, r);
  tree[v] = f(tree[2 * v], tree[2 * v + 1]);
}
Node query(int lq, int rq, int v = 1, int l = 0, int r = n) {
  if (lq >= r || rq <= l) return Node();
  if (lq <= l && r <= rq) return tree[v];
  int md = (l + r) / 2;
  return f(query(lq, rq, 2 * v, l, md), query(lq, rq, 2 * v + 1, md, r));
}
void solve() {
  int q;
  cin >> n >> q;
  cin >> s;
  build();
  for (int k = 0; k < q; k++) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    Node mat = query(a, b + 1);
    int res = mat[0][4];
    if (res >= Inf) res = -1;
    cout << res << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(9);
  cout << fixed;
  solve();
  return 0;
}

### 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 Inf = 1e9 + 7;
const int N = 2e5 + 7;
int n;
string s;
struct Node {
  int m[5][5];
  int *operator[](int x) { return m[x]; }
  Node() : Node('?') {}
  Node(char x) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) m[i][j] = (i != j ? Inf : 0);
    switch (x) {
      case '2':
        m[0][0] = 1;
        m[0][1] = 0;
        break;
      case '0':
        m[1][1] = 1;
        m[1][2] = 0;
        break;
      case '1':
        m[2][2] = 1;
        m[2][3] = 0;
        break;
      case '7':
        m[3][3] = 1;
        m[3][4] = 0;
        break;
      case '6':
        m[3][3] = 1;
        m[4][4] = 1;
      default:
        break;
    }
  }
};
Node tree[4 * N];
Node f(const Node &a, const Node &b) {
  Node res;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      res[i][j] = Inf;
      for (int k = 0; k < 5; k++)
        res[i][j] = min(res[i][j], a.m[i][k] + b.m[k][j]);
    }
  }
  return res;
}
void build(int v = 1, int l = 0, int r = n) {
  if (l + 1 == r) {
    tree[v] = Node(s[l]);
    return;
  }
  int md = (l + r) / 2;
  build(2 * v, l, md);
  build(2 * v + 1, md, r);
  tree[v] = f(tree[2 * v], tree[2 * v + 1]);
}
Node query(int lq, int rq, int v = 1, int l = 0, int r = n) {
  if (lq >= r || rq <= l) return Node();
  if (lq <= l && r <= rq) return tree[v];
  int md = (l + r) / 2;
  return f(query(lq, rq, 2 * v, l, md), query(lq, rq, 2 * v + 1, md, r));
}
void solve() {
  int q;
  cin >> n >> q;
  cin >> s;
  build();
  for (int k = 0; k < q; k++) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    Node mat = query(a, b + 1);
    int res = mat[0][4];
    if (res >= Inf) res = -1;
    cout << res << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(9);
  cout << fixed;
  solve();
  return 0;
}

```