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 = 5000;
vector<pair<int, int>> merges;
vector<int> arr;
int cnt;
struct data {
  vector<int> v;
  map<pair<int, int>, int> m;
} st[N * 4];
void combine(int root) {
  vector<int> &v1 = st[2 * root].v;
  vector<int> &v2 = st[2 * root + 1].v;
  vector<int> &v = st[root].v;
  v.resize((int)v1.size() + (int)v2.size());
  merge(v1.begin(), v1.end(), v2.begin(), v2.end(), v.begin());
}
void build(int root, int sl, int sr) {
  if (sl == sr) {
    st[root].v.push_back(arr[sl]);
    return;
  }
  int mid = (sl + sr) / 2;
  build(2 * root, sl, mid);
  build(2 * root + 1, mid + 1, sr);
  combine(root);
}
int query(int root, int l, int r) {
  vector<int> &v1 = st[root].v;
  int x = lower_bound(v1.begin(), v1.end(), l) - v1.begin();
  int y = upper_bound(v1.begin(), v1.end(), r) - v1.begin();
  if (y == x + 1) {
    return v1[x];
  }
  if (x == (int)v1.size() || (x >= y)) {
    return -1;
  }
  map<pair<int, int>, int> &m = st[root].m;
  if (m.find({x, y}) == m.end()) {
    int s1 = query(2 * root, l, r);
    int s2 = query(2 * root + 1, l, r);
    if (s1 != -1 && s2 != -1) {
      merges.push_back({s1, s2});
      m[{x, y}] = cnt + 1;
      cnt += 1;
    } else {
      m[{x, y}] = max(s1, s2);
    }
  }
  return m[{x, y}];
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  cnt = n;
  arr.clear();
  merges.clear();
  arr.resize(n + 1, -1);
  for (int i = 0; i < n; ++i) {
    int a;
    scanf("%d", &a);
    arr[a] = i + 1;
  }
  build(1, 1, n);
  vector<int> res;
  while (q) {
    q -= 1;
    int l, r;
    scanf("%d%d", &l, &r);
    res.push_back(query(1, l, r));
  }
  printf("%d\n", cnt);
  for (auto x : merges) {
    printf("%d %d\n", x.first, x.second);
  }
  for (int x : res) {
    printf("%d ", x);
  }
  printf("\n");
}