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;
const int N = 4100;
vector<pair<int, int>> ans;
map<pair<int, int>, int> dp[N];
int n, q, l, r, s, a[N], m[N], le[N], ri[N], k[66000], mx[2200020];
map<pair<int, int>, int> f(int lo, int hi) {
  if (lo == hi) return {{{lo, hi}, m[lo]}};
  if (lo < hi) {
    int mi = (lo + hi) / 2;
    auto m0 = f(lo, mi);
    auto m1 = f(mi + 1, hi);
    vector<int> v;
    for (int i = 1; i <= n; i++)
      if (lo <= a[i] && a[i] <= hi) v.push_back(a[i]);
    map<pair<int, int>, int> res;
    for (int i = 0; i < v.size(); i++) {
      le[0] = le[1] = 0;
      for (int j = i; j < v.size(); j++) {
        int t = v[j] > mi;
        if (!le[t]) le[t] = v[j];
        ri[t] = v[j];
        int k0 = m0[{le[0], ri[0]}];
        int k1 = m1[{le[1], ri[1]}];
        if (!res[{le[0], ri[0]}]) res[{le[0], ri[0]}] = k0;
        if (!res[{le[1], ri[1]}]) res[{le[1], ri[1]}] = k1;
        if (le[0] && le[1]) {
          ans.emplace_back(k0, k1);
          res[{v[i], v[j]}] = n + ans.size();
        }
      }
      res[{v[i], v[i]}] = m[v[i]];
    }
    return res;
  }
}
int main() {
  scanf("%d%d", &n, &q);
  for (; s * s < q; s++)
    ;
  for (int i = 1; i <= n; i++) scanf("%d", a + i), m[a[i]] = i, mx[i] = a[i];
  for (int i = 1, j = 0; i <= n; i += s, j++) dp[j] = f(i, i + s - 1);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &l, &r);
    for (int j = 0; j < n; j++) le[j] = N - 1, ri[j] = 0;
    for (int j = l; j <= r; j++) {
      int t = (a[j] - 1) / s;
      le[t] = min(le[t], m[a[j]]), ri[t] = max(ri[t], m[a[j]]);
    }
    for (int j = 0, nxt, prv = 0; j <= n / s; j++) {
      nxt = dp[j][{a[le[j]], a[ri[j]]}];
      if (prv && nxt) ans.emplace_back(prv, nxt), nxt = n + ans.size();
      if (nxt) k[i] = prv = nxt;
    }
  }
  printf("%d\n", n + ans.size());
  for (int i = 0; i < ans.size(); i++) {
    if (mx[ans[i].first] > mx[ans[i].second]) swap(ans[i].first, ans[i].second);
    printf("%d %d\n", ans[i].first, ans[i].second);
    mx[n + 1 + i] = mx[ans[i].second];
  }
  for (int i = 0; i < q; i++) printf("%d ", k[i]);
}