#include <bits/stdc++.h>
using namespace std;
template <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
static void redirect(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
}
struct node {
  int dp[5][5];
  node() {
    memset(dp, -1, sizeof(dp));
    for (int i = 0; i <= 4; i++) dp[i][i] = 0;
  }
};
static node join(const node &a, const node &b) {
  node out;
  for (int i = 0; i <= 4; i++)
    for (int j = i; j <= 4; j++) {
      int best = INT_MAX;
      for (int k = i; k <= j; k++)
        if (a.dp[i][k] != -1 && b.dp[k][j] != -1)
          best = min(best, a.dp[i][k] + b.dp[k][j]);
      if (best != INT_MAX)
        out.dp[i][j] = best;
      else
        out.dp[i][j] = -1;
    }
  return out;
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  int N, Q;
  string s;
  cin >> N >> Q >> s;
  int ofs = 1;
  while (ofs < N) ofs *= 2;
  vector<node> nodes(2 * ofs);
  for (int i = 0; i < N; i++) {
    node &cur = nodes[i + ofs];
    switch (s[i]) {
      case '2':
        cur.dp[0][1] = 0;
        cur.dp[0][0] = 1;
        break;
      case '0':
        cur.dp[1][2] = 0;
        cur.dp[1][1] = 1;
        break;
      case '1':
        cur.dp[2][3] = 0;
        cur.dp[2][2] = 1;
        break;
      case '6':
        cur.dp[3][3] = 1;
        cur.dp[4][4] = 1;
        break;
      case '7':
        cur.dp[3][4] = 0;
        cur.dp[3][3] = 1;
        break;
    }
  }
  for (int i = ofs - 1; i > 0; i--)
    nodes[i] = join(nodes[2 * i], nodes[2 * i + 1]);
  for (int i = 0; i < Q; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    a += ofs;
    b += ofs;
    node lft, rgt;
    while (a < b) {
      if (a & 1) lft = join(lft, nodes[a++]);
      if (b & 1) rgt = join(nodes[--b], rgt);
      a >>= 1;
      b >>= 1;
    }
    node ans = join(lft, rgt);
    cout << ans.dp[0][4] << '\n';
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
static void redirect(int argc, const char **argv) {
  ios::sync_with_stdio(false);
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
}
struct node {
  int dp[5][5];
  node() {
    memset(dp, -1, sizeof(dp));
    for (int i = 0; i <= 4; i++) dp[i][i] = 0;
  }
};
static node join(const node &a, const node &b) {
  node out;
  for (int i = 0; i <= 4; i++)
    for (int j = i; j <= 4; j++) {
      int best = INT_MAX;
      for (int k = i; k <= j; k++)
        if (a.dp[i][k] != -1 && b.dp[k][j] != -1)
          best = min(best, a.dp[i][k] + b.dp[k][j]);
      if (best != INT_MAX)
        out.dp[i][j] = best;
      else
        out.dp[i][j] = -1;
    }
  return out;
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  int N, Q;
  string s;
  cin >> N >> Q >> s;
  int ofs = 1;
  while (ofs < N) ofs *= 2;
  vector<node> nodes(2 * ofs);
  for (int i = 0; i < N; i++) {
    node &cur = nodes[i + ofs];
    switch (s[i]) {
      case '2':
        cur.dp[0][1] = 0;
        cur.dp[0][0] = 1;
        break;
      case '0':
        cur.dp[1][2] = 0;
        cur.dp[1][1] = 1;
        break;
      case '1':
        cur.dp[2][3] = 0;
        cur.dp[2][2] = 1;
        break;
      case '6':
        cur.dp[3][3] = 1;
        cur.dp[4][4] = 1;
        break;
      case '7':
        cur.dp[3][4] = 0;
        cur.dp[3][3] = 1;
        break;
    }
  }
  for (int i = ofs - 1; i > 0; i--)
    nodes[i] = join(nodes[2 * i], nodes[2 * i + 1]);
  for (int i = 0; i < Q; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    a += ofs;
    b += ofs;
    node lft, rgt;
    while (a < b) {
      if (a & 1) lft = join(lft, nodes[a++]);
      if (b & 1) rgt = join(nodes[--b], rgt);
      a >>= 1;
      b >>= 1;
    }
    node ans = join(lft, rgt);
    cout << ans.dp[0][4] << '\n';
  }
  return 0;
}

```