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;
using ll = long long;
using vi = vector<int>;
using ld = long double;
using pii = pair<int, int>;
int S;
vector<pii> ans;
pii getBounds(set<int> s, int a, int b) {
  auto l = s.lower_bound(a);
  auto r = prev(s.upper_bound(b));
  return {*l, *r};
}
struct node {
  int lo, hi, mid;
  node *left, *right;
  map<pii, int> dp;
  set<int> idx;
  node(int a, int b, vi &ids) {
    lo = a, hi = b;
    mid = (lo + hi) / 2;
    if (lo != hi) {
      left = new node(lo, mid, ids);
      right = new node(mid + 1, hi, ids);
      idx.insert(left->idx.begin(), left->idx.end());
      idx.insert(right->idx.begin(), right->idx.end());
    } else {
      dp[{ids[hi], ids[hi]}] = ids[hi];
      idx.insert(ids[hi]);
    }
  }
  int query(int a, int b) {
    auto lb = idx.lower_bound(a);
    auto rb = idx.upper_bound(b);
    if (lb == idx.end() || *lb > b) return -1;
    a = *lb;
    b = *prev(rb);
    if (dp.count({a, b})) return dp[{a, b}];
    int x = -1, y = -1;
    x = left->query(a, b);
    y = right->query(a, b);
    if (x == -1 || y == -1) return dp[{a, b}] = max(x, y);
    ans.emplace_back(x, y);
    return dp[{a, b}] = ++S;
  }
};
void solve() {
  int n, q;
  cin >> n >> q;
  vi a(n), ids(n);
  for (int i = 0; i < n; i++) cin >> a[i], ids[--a[i]] = i + 1;
  S = n;
  node *root = new node(0, n - 1, ids);
  vi ansK;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    ansK.push_back(root->query(l, r));
  }
  cout << S << '\n';
  for (auto it : ans) cout << it.first << " " << it.second << '\n';
  for (int i : ansK) cout << i << " ";
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}