#include <bits/stdc++.h>
using namespace std;
struct st {
  int dp[5][5];
  st(void) {
    for (int i = 0; i <= 4; ++i)
      for (int j = 0; j <= 4; ++j) dp[i][j] = 1e9;
  };
};
st operator+(st a, st b) {
  st c;
  for (int i = 0; i <= 4; ++i)
    for (int k = i; k <= 4; ++k)
      for (int j = k; j <= 4; ++j)
        c.dp[i][j] = min(c.dp[i][j], a.dp[i][k] + b.dp[k][j]);
  return c;
};
st t[1 << 21];
char s[1 << 20];
int v[1 << 20];
void build(int p, int u, int node) {
  if (p == u) {
    for (int i = 0; i <= 4; ++i) t[node].dp[i][i] = 0;
    if (!v[p]) return;
    if (v[p] <= 3)
      t[node].dp[v[p] - 1][v[p] - 1] = 1, t[node].dp[v[p] - 1][v[p]] = 0;
    else if (v[p] == 4)
      t[node].dp[3][4] = 0;
    else
      t[node].dp[3][3] = t[node].dp[4][4] = 1;
  } else {
    int m = (p + u) / 2;
    build(p, m, node << 1);
    build(m + 1, u, node << 1 | 1);
    t[node] = t[node << 1] + t[node << 1 | 1];
  }
}
st query(int p, int u, int l, int r, int node) {
  st ans;
  if (l <= p && u <= r)
    ans = t[node];
  else {
    int m = (p + u) / 2;
    if (l <= m && m + 1 <= r)
      ans = query(p, m, l, r, node << 1) + query(m + 1, u, l, r, node << 1 | 1);
    else if (l <= m)
      ans = query(p, m, l, r, node << 1);
    else
      ans = query(m + 1, u, l, r, node << 1 | 1);
  }
  return ans;
}
int main(void) {
  srand(time(0));
  cout << fixed << setprecision(7);
  cerr << fixed << setprecision(7);
  int n, q;
  cin >> n >> q;
  cin >> (s + 1);
  for (int i = 1; i <= n; ++i)
    if (s[i] == '2')
      v[i] = 1;
    else if (s[i] == '0')
      v[i] = 2;
    else if (s[i] == '1')
      v[i] = 3;
    else if (s[i] == '7')
      v[i] = 4;
    else if (s[i] == '6')
      v[i] = 5;
  build(1, n, 1);
  while (q--) {
    int l, r;
    cin >> l >> r;
    auto brut = [&] {
      static int dp[6];
      for (int i = 0; i < 5; ++i) dp[i] = 1e9;
      dp[0] = 0;
      for (int i = l; i <= r; ++i)
        if (v[i]) {
          if (v[i] == 5)
            ++dp[3], ++dp[4];
          else {
            dp[v[i]] = min(dp[v[i]], dp[v[i] - 1]);
            ++dp[v[i] - 1];
          }
        }
      if (dp[4] > 1e7)
        return -1;
      else
        return dp[4];
    };
    int ans = query(1, n, l, r, 1).dp[0][4];
    if (ans > 1e7) ans = -1;
    cout << ans << '\n';
  }
  cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms"
       << '\n';
  return 0;
}

### 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;
struct st {
  int dp[5][5];
  st(void) {
    for (int i = 0; i <= 4; ++i)
      for (int j = 0; j <= 4; ++j) dp[i][j] = 1e9;
  };
};
st operator+(st a, st b) {
  st c;
  for (int i = 0; i <= 4; ++i)
    for (int k = i; k <= 4; ++k)
      for (int j = k; j <= 4; ++j)
        c.dp[i][j] = min(c.dp[i][j], a.dp[i][k] + b.dp[k][j]);
  return c;
};
st t[1 << 21];
char s[1 << 20];
int v[1 << 20];
void build(int p, int u, int node) {
  if (p == u) {
    for (int i = 0; i <= 4; ++i) t[node].dp[i][i] = 0;
    if (!v[p]) return;
    if (v[p] <= 3)
      t[node].dp[v[p] - 1][v[p] - 1] = 1, t[node].dp[v[p] - 1][v[p]] = 0;
    else if (v[p] == 4)
      t[node].dp[3][4] = 0;
    else
      t[node].dp[3][3] = t[node].dp[4][4] = 1;
  } else {
    int m = (p + u) / 2;
    build(p, m, node << 1);
    build(m + 1, u, node << 1 | 1);
    t[node] = t[node << 1] + t[node << 1 | 1];
  }
}
st query(int p, int u, int l, int r, int node) {
  st ans;
  if (l <= p && u <= r)
    ans = t[node];
  else {
    int m = (p + u) / 2;
    if (l <= m && m + 1 <= r)
      ans = query(p, m, l, r, node << 1) + query(m + 1, u, l, r, node << 1 | 1);
    else if (l <= m)
      ans = query(p, m, l, r, node << 1);
    else
      ans = query(m + 1, u, l, r, node << 1 | 1);
  }
  return ans;
}
int main(void) {
  srand(time(0));
  cout << fixed << setprecision(7);
  cerr << fixed << setprecision(7);
  int n, q;
  cin >> n >> q;
  cin >> (s + 1);
  for (int i = 1; i <= n; ++i)
    if (s[i] == '2')
      v[i] = 1;
    else if (s[i] == '0')
      v[i] = 2;
    else if (s[i] == '1')
      v[i] = 3;
    else if (s[i] == '7')
      v[i] = 4;
    else if (s[i] == '6')
      v[i] = 5;
  build(1, n, 1);
  while (q--) {
    int l, r;
    cin >> l >> r;
    auto brut = [&] {
      static int dp[6];
      for (int i = 0; i < 5; ++i) dp[i] = 1e9;
      dp[0] = 0;
      for (int i = l; i <= r; ++i)
        if (v[i]) {
          if (v[i] == 5)
            ++dp[3], ++dp[4];
          else {
            dp[v[i]] = min(dp[v[i]], dp[v[i] - 1]);
            ++dp[v[i] - 1];
          }
        }
      if (dp[4] > 1e7)
        return -1;
      else
        return dp[4];
    };
    int ans = query(1, n, l, r, 1).dp[0][4];
    if (ans > 1e7) ans = -1;
    cout << ans << '\n';
  }
  cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms"
       << '\n';
  return 0;
}

```