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;
struct node {
  vector<int> num;
  map<pair<int, int>, int> Hash;
} t[100005 << 2];
pair<int, int> vis[2200005];
int n, Q, cnt;
int a[100005], pos[100005], ans[100005];
void build(int now, int l, int r) {
  for (int i = l; i <= r; i++) t[now].num.push_back(pos[i]);
  sort(t[now].num.begin(), t[now].num.end());
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(now << 1, l, mid), build(now << 1 | 1, mid + 1, r);
}
int query(int now, int l, int r) {
  int left =
      lower_bound(t[now].num.begin(), t[now].num.end(), l) - t[now].num.begin();
  int right = upper_bound(t[now].num.begin(), t[now].num.end(), r) -
              t[now].num.begin() - 1;
  if (right < left) return 0;
  if (left == right) return t[now].num[left];
  int pos = t[now].Hash[make_pair(left, right)];
  if (pos) return pos;
  int lson = query(now << 1, l, r), rson = query(now << 1 | 1, l, r);
  if (!lson || !rson) return t[now].Hash[make_pair(left, right)] = lson | rson;
  vis[++cnt] = make_pair(lson, rson);
  return t[now].Hash[make_pair(left, right)] = cnt;
}
int main() {
  scanf("%d %d", &n, &Q);
  cnt = n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), pos[a[i]] = i;
  build(1, 1, n);
  for (int i = 1, l, r; i <= Q; i++) {
    scanf("%d %d", &l, &r);
    ans[i] = query(1, l, r);
  }
  printf("%d\n", cnt);
  for (int i = n + 1; i <= cnt; i++)
    printf("%d %d\n", vis[i].first, vis[i].second);
  for (int i = 1; i <= Q; i++) printf("%d ", ans[i]);
  return 0;
}