Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, q, a[(1 << 12 | 5)], loc[(1 << 12 | 5)], cnt;
int l[(1 << 16 | 5)], r[(1 << 16 | 5)], ans[(1 << 16 | 5)];
int ls[(1 << 12 | 5) << 2], rs[(1 << 12 | 5) << 2];
;
vector<vector<int> > id[(1 << 12 | 5) << 2];
vector<pair<int, int> > b[(1 << 12 | 5) << 4], fk;
void Merge(int first, int second) {
  fk.push_back(make_pair(first, second));
  cnt++;
}
int tot, root[(1 << 12 | 5)];
int qry(int first, int l, int r) {
  int L = lower_bound(b[first].begin(), b[first].end(), make_pair(l, 0)) -
          b[first].begin();
  int R = lower_bound(b[first].begin(), b[first].end(), make_pair(r + 1, 0)) -
          b[first].begin() - 1;
  if (L > R) return 0;
  return id[first][L][R];
}
void build(int &first, int l, int r, int st) {
  first = ++tot;
  assert(!l);
  if (l == r) {
    b[first].push_back(make_pair(loc[st + l], st + l));
    id[first].resize(1);
    id[first][0].resize(1);
    id[first][0][0] = loc[st + l];
    return;
  }
  int mid = (l + r) >> 1;
  build(ls[first], l, mid, st);
  build(rs[first], 0, r - mid - 1, st + mid + 1 - l);
  b[first].resize(r + 1);
  id[first].resize(r + 1);
  for (auto &i : id[first]) i.resize(r + 1);
  merge(b[ls[first]].begin(), b[ls[first]].end(), b[rs[first]].begin(),
        b[rs[first]].end(), b[first].begin());
  for (int i = l; i <= r; ++i)
    for (int j = i; j <= r; ++j) {
      int X = qry(ls[first], b[first][i].first, b[first][j].first),
          Y = qry(rs[first], b[first][i].first, b[first][j].first);
      if (X && Y) {
        Merge(X, Y);
        id[first][i][j] = cnt;
      } else if (X)
        id[first][i][j] = X;
      else if (Y)
        id[first][i][j] = Y;
    }
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    loc[a[i]] = i;
  }
  cnt = n;
  int B = min(n, (int)sqrt(q));
  int T = (n - 1) / B + 1;
  for (int i = 1; i <= T; ++i) {
    int l = (i - 1) * B + 1, r = min(n, i * B);
    build(root[i], 0, r - l, l);
  }
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d", &l[i], &r[i]);
    int lst = 0;
    for (int j = 1; j <= T; ++j) {
      int ID = qry(root[j], l[i], r[i]);
      if (lst && ID) {
        Merge(lst, ID);
        lst = cnt;
      } else if (ID)
        lst = ID;
    }
    ans[i] = lst;
  }
  printf("%d\n", n + (int)fk.size());
  for (auto i : fk) printf("%d %d\n", i.first, i.second);
  for (int i = 1; i <= q; ++i) printf("%d ", ans[i]);
  return 0;
}