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>
const int N = 4096;
const int maxn = 3e6;
const int M = 256;
using namespace std;
struct Ans {
  int x, y;
} s[maxn + 5];
struct block {
  int mp[M + 2][M + 2], id[N + 1];
} b[16 + 1];
int n, q, a[N + 5], t[N + 5], L[N + 5], R[N + 5], blo[N + 5], bs, ans, sm, cnt,
    anss[maxn + 5], f[M * 2 + 5][M + 2][M + 2];
int merge(int x, int y) {
  if (!x || !y) return x + y;
  s[++ans] = (Ans){x, y};
  return ++sm;
}
void solve(int k, int l, int r, vector<int> c) {
  if (l == r) {
    f[k][1][1] = t[c[1]];
    return;
  }
  int mid = l + r >> 1;
  vector<int> cl(1), cr(1);
  for (int i = 1; i < c.size(); i++)
    if (c[i] <= mid)
      cl.push_back(c[i]);
    else
      cr.push_back(c[i]);
  solve(k << 1, l, mid, cl);
  solve(k << 1 | 1, mid + 1, r, cr);
  int L = 1, R = 1;
  for (int i = 1; i < c.size(); c[i] <= mid ? L++ : R++, i++) {
    int ll = L - 1, rr = R - 1;
    c[i] <= mid ? ll++ : rr++;
    for (int j = i; j < c.size(); j++, c[j] <= mid ? ll++ : rr++)
      f[k][i][j] = merge(f[k << 1][L][ll], f[k << 1 | 1][R][rr]);
  }
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), t[a[i]] = i;
  sm = n;
  bs = M;
  for (int i = 1; i <= n; i++) {
    blo[i] = (i - 1) / bs + 1;
    if (!L[blo[i]]) L[blo[i]] = i;
    R[blo[i]] = i;
  }
  for (int x = 1; x <= blo[n]; x++) {
    vector<int> c(1);
    for (int i = 1; i <= n; i++)
      if (a[i] >= L[x] && a[i] <= R[x]) {
        c.push_back(a[i]);
        b[x].id[i] = b[x].id[i - 1] + 1;
      } else
        b[x].id[i] = b[x].id[i - 1];
    solve(1, L[x], R[x], c);
    for (int i = 1; i < c.size(); i++)
      for (int j = 1; j < c.size(); j++) b[x].mp[i][j] = f[1][i][j];
  }
  int l, r;
  while (q--) {
    scanf("%d%d", &l, &r);
    cnt++;
    for (int x = 1; x <= blo[n]; x++)
      anss[cnt] = merge(anss[cnt], b[x].mp[b[x].id[l - 1] + 1][b[x].id[r]]);
  }
  cout << sm << endl;
  for (int i = 1; i <= ans; i++) printf("%d %d\n", s[i].x, s[i].y);
  for (int i = 1; i <= cnt; i++) printf("%d ", anss[i]);
  return 0;
}