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 opt[2200010][2], ocnt = 0;
int merge(int x, int y) {
  if (!x || !y) return x | y;
  ++ocnt;
  opt[ocnt][0] = x;
  opt[ocnt][1] = y;
  return ocnt;
}
const int B = 1 << 8;
int buf[(1 << 9) + 1][1 << 8][1 << 8];
class block {
 public:
  int st[1 << 8][1 << 8], pos[1 << 8], siz;
  vector<int> seq;
  void build(int p, int l, int r, vector<int> ss) {
    if (l == r) {
      buf[p][0][0] = pos[l] + 1;
      return;
    }
    int mid = (l + r) >> 1;
    vector<int> vl, vr;
    for (int i = 0; i < ss.size(); i++) {
      if (ss[i] <= mid)
        vl.push_back(ss[i]);
      else
        vr.push_back(ss[i]);
    }
    build(p << 1, l, mid, vl);
    build((p << 1) | 1, mid + 1, r, vr);
    int l1 = 0, r1 = 0, l2 = 0, r2 = 0;
    for (int i = 0; i <= r - l; i++) {
      r1 = l1 - 1;
      r2 = l2 - 1;
      for (int j = i; j <= r - l; j++) {
        if (ss[j] <= mid)
          r1++;
        else
          r2++;
        buf[p][i][j] = merge(l1 <= r1 ? buf[p << 1][l1][r1] : 0,
                             l2 <= r2 ? buf[(p << 1) | 1][l2][r2] : 0);
      }
      if (ss[i] <= mid)
        l1++;
      else
        l2++;
    }
  }
  int query(int l, int r) {
    l = lower_bound(seq.begin(), seq.end(), l) - seq.begin();
    r = upper_bound(seq.begin(), seq.end(), r) - seq.begin() - 1;
    if (l > r) return 0;
    return st[l][r];
  }
} bl[1 << 4];
int n, Q, a[1 << 12], ans[1 << 16];
int main() {
  scanf("%d%d", &n, &Q);
  ocnt = n;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), a[i]--;
  for (int i = 0; i < n; i += B) {
    int l = i, r = min(i + B, n) - 1;
    int p = i / B;
    vector<int> ss;
    for (int j = 0; j < n; j++) {
      if (a[j] >= l && a[j] <= r) {
        bl[p].pos[a[j] - l] = j;
        ss.push_back(a[j] - l);
        bl[p].seq.push_back(j);
      }
    }
    bl[p].build(1, 0, r - l, ss);
    memcpy(bl[p].st, buf[1], sizeof(buf[1]));
  }
  for (int i = 0, l, r; i < Q; i++) {
    scanf("%d%d", &l, &r);
    l--, r--;
    int cur = 0;
    for (int j = 0; j * B < n; j++) cur = merge(cur, bl[j].query(l, r));
    ans[i] = cur;
  }
  printf("%d\n", ocnt);
  for (int i = n + 1; i <= ocnt; i++) printf("%d %d\n", opt[i][0], opt[i][1]);
  for (int i = 0; i < Q; i++) printf("%d ", ans[i]);
  return 0;
}