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;
vector<int> blocks[5005];
map<pair<int, int>, int> pos[5005];
vector<pair<int, int> > ops;
int a[5005], id[5005];
int ret[100005];
struct Partition {
  map<pair<int, int>, int> ranges;
  Partition() {}
};
int cnt;
Partition construct(int low, int high, vector<int>& order) {
  if (low > high) {
    return Partition();
  }
  if (low == high) {
    Partition ps;
    ps.ranges[make_pair(0, 0)] = id[low];
    return ps;
  }
  int mid = (low + high) >> 1;
  vector<int> la, ra;
  for (int x : order) {
    if (x <= mid) {
      la.push_back(x);
    } else {
      ra.push_back(x);
    }
  }
  auto ls = construct(low, mid, la);
  auto rs = construct(mid + 1, high, ra);
  Partition ps;
  int i = 0;
  int lowx = 0, lowy = 0, highx = 0, highy = 0;
  while (i < order.size()) {
    if (i > 0) {
      if (order[i - 1] <= mid) {
        lowx++;
      } else {
        highx++;
      }
    }
    lowy = lowx - 1;
    highy = highx - 1;
    for (int j = i; j < order.size(); j++) {
      if (order[j] <= mid) {
        lowy++;
      } else {
        highy++;
      }
      int lindex = ls.ranges[make_pair(lowx, lowy)];
      int rindex = rs.ranges[make_pair(highx, highy)];
      if (lowx <= lowy && highx <= highy) {
        assert(lindex > 0 && rindex > 0);
        ps.ranges[make_pair(i, j)] = ++cnt;
        ops.push_back(make_pair(lindex, rindex));
      } else if (lowx <= lowy) {
        assert(lindex > 0);
        ps.ranges[make_pair(i, j)] = lindex;
      } else {
        if (rindex == 0) {
        }
        assert(rindex > 0);
        ps.ranges[make_pair(i, j)] = rindex;
      }
    }
    i++;
  }
  return ps;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    id[a[i]] = i;
  }
  int S = sqrt(q);
  cnt = n;
  int nblocks = 0;
  for (int low = 1; low <= n; low += S, nblocks++) {
    int high = min(n, low + S - 1);
    for (int i = 1; i <= n; i++)
      if (a[i] >= low && a[i] <= high) {
        blocks[nblocks].push_back(a[i]);
      }
    auto ps = construct(low, high, blocks[nblocks]);
    for (int i = 0; i < blocks[nblocks].size(); i++) {
      for (int j = i; j < blocks[nblocks].size(); j++) {
        pos[nblocks][make_pair(i, j)] = ps.ranges[make_pair(i, j)];
      }
    }
  }
  for (int it = 0; it < q; it++) {
    int l, r;
    scanf("%d %d", &l, &r);
    ret[it] = -1;
    vector<pair<int, int> > ranges;
    ranges.resize(nblocks);
    for (int i = 0; i < nblocks; i++) {
      ranges[i] = make_pair(0, -1);
    }
    for (int i = 1; i <= r; i++) {
      int block = (a[i] - 1) / S;
      if (i < l) {
        ranges[block].first++;
        ranges[block].second++;
      } else {
        ranges[block].second++;
      }
    }
    for (int i = 0; i < nblocks; i++)
      if (ranges[i].first <= ranges[i].second) {
        int lindex = pos[i][make_pair(ranges[i].first, ranges[i].second)];
        if (ret[it] < 0) {
          ret[it] = lindex;
        } else {
          assert(ret[it] > 0 && lindex > 0);
          ops.push_back(make_pair(ret[it], lindex));
          ret[it] = ++cnt;
        }
      }
  }
  printf("%d\n", cnt);
  for (auto op : ops) {
    printf("%d %d\n", op.first, op.second);
  }
  for (int i = 0; i < q; i++) {
    printf("%d ", ret[i]);
  }
  printf("\n");
  return 0;
}