#include <bits/stdc++.h>
using namespace std;
template <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  int l = v.size();
  for (int i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << endl;
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
const long double PI = 3.1415926535897932384626433832795;
const int N = 1 << 19;
const int M = 5;
const int INF = 1e8;
struct node {
  int mcost[M][M];
  node() {
    for (int i = 0; i < M; i++) {
      for (int j = 0; j < M; j++) {
        mcost[i][j] = (i == j) ? 0 : INF;
      }
    }
  }
  void assign(char d) {
    if (d == '2') {
      mcost[0][0] = 1;
      mcost[0][1] = 0;
    } else if (d == '0') {
      mcost[1][1] = 1;
      mcost[1][2] = 0;
    } else if (d == '1') {
      mcost[2][2] = 1;
      mcost[2][3] = 0;
    } else if (d == '7') {
      mcost[3][3] = 1;
      mcost[3][4] = 0;
    } else if (d == '6') {
      mcost[3][3] = 1;
      mcost[4][4] = 1;
    }
  }
  void combine(node& left, node& right) {
    for (int i = 0; i < M; i++) {
      for (int j = i; j < M; j++) {
        mcost[i][j] = min(left.mcost[i][i] + right.mcost[i][j],
                          left.mcost[i][j] + right.mcost[j][j]);
        for (int k = i + 1; k < j; k++) {
          mcost[i][j] = min(mcost[i][j], left.mcost[i][k] + right.mcost[k][j]);
        }
      }
    }
  }
};
int n, q;
string s;
node tree[2 * N];
void build(int id = 1, int l = 0, int r = n) {
  if (l + 1 == r) {
    tree[id].assign(s[l]);
    return;
  }
  int left = id << 1, right = left + 1, mid = (l + r) >> 1;
  build(left, l, mid);
  build(right, mid, r);
  tree[id].combine(tree[left], tree[right]);
  return;
}
node query(int x, int y, int id = 1, int l = 0, int r = n) {
  if (x >= r or l >= y) return node();
  if (x <= l && r <= y) return tree[id];
  int left = id << 1, right = left + 1, mid = (l + r) >> 1;
  node lft = query(x, y, left, l, mid), rgt = query(x, y, right, mid, r);
  node ret = node();
  ret.combine(lft, rgt);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> q;
  cin >> s;
  build();
  while (q--) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    node ans = query(a, b + 1);
    if (ans.mcost[0][M - 1] >= INF)
      cout << -1 << endl;
    else
      cout << ans.mcost[0][M - 1] << endl;
  }
  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;
template <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  int l = v.size();
  for (int i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << endl;
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
const long double PI = 3.1415926535897932384626433832795;
const int N = 1 << 19;
const int M = 5;
const int INF = 1e8;
struct node {
  int mcost[M][M];
  node() {
    for (int i = 0; i < M; i++) {
      for (int j = 0; j < M; j++) {
        mcost[i][j] = (i == j) ? 0 : INF;
      }
    }
  }
  void assign(char d) {
    if (d == '2') {
      mcost[0][0] = 1;
      mcost[0][1] = 0;
    } else if (d == '0') {
      mcost[1][1] = 1;
      mcost[1][2] = 0;
    } else if (d == '1') {
      mcost[2][2] = 1;
      mcost[2][3] = 0;
    } else if (d == '7') {
      mcost[3][3] = 1;
      mcost[3][4] = 0;
    } else if (d == '6') {
      mcost[3][3] = 1;
      mcost[4][4] = 1;
    }
  }
  void combine(node& left, node& right) {
    for (int i = 0; i < M; i++) {
      for (int j = i; j < M; j++) {
        mcost[i][j] = min(left.mcost[i][i] + right.mcost[i][j],
                          left.mcost[i][j] + right.mcost[j][j]);
        for (int k = i + 1; k < j; k++) {
          mcost[i][j] = min(mcost[i][j], left.mcost[i][k] + right.mcost[k][j]);
        }
      }
    }
  }
};
int n, q;
string s;
node tree[2 * N];
void build(int id = 1, int l = 0, int r = n) {
  if (l + 1 == r) {
    tree[id].assign(s[l]);
    return;
  }
  int left = id << 1, right = left + 1, mid = (l + r) >> 1;
  build(left, l, mid);
  build(right, mid, r);
  tree[id].combine(tree[left], tree[right]);
  return;
}
node query(int x, int y, int id = 1, int l = 0, int r = n) {
  if (x >= r or l >= y) return node();
  if (x <= l && r <= y) return tree[id];
  int left = id << 1, right = left + 1, mid = (l + r) >> 1;
  node lft = query(x, y, left, l, mid), rgt = query(x, y, right, mid, r);
  node ret = node();
  ret.combine(lft, rgt);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> q;
  cin >> s;
  build();
  while (q--) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    node ans = query(a, b + 1);
    if (ans.mcost[0][M - 1] >= INF)
      cout << -1 << endl;
    else
      cout << ans.mcost[0][M - 1] << endl;
  }
  return 0;
}

```