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;
inline int Read() {
  int x(0), f(1);
  char c(getchar());
  while (c < '0' || '9' < c) {
    if (c == '-') {
      f = -1;
    }
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return f * x;
}
void Write(int x) {
  if (x < 0) {
    printf("-");
    x = -x;
  }
  if (9 < x) {
    Write(x / 10);
  }
  putchar(x % 10 + '0');
}
inline void Writeln(int x) {
  Write(x);
  puts("");
}
const int MAXN = 1e5 + 5;
const int MAXQ = 2e6 + 2e5 + 5;
int n, q;
int arr[MAXN];
int app[MAXN];
struct Range {
  int left, right;
  Range(int l = 0, int r = 0) {
    left = l;
    right = r;
  }
} use[MAXQ];
int answer[MAXN];
int cnt = 0;
struct SegmentTree {
  vector<int> num;
  map<pair<int, int>, int> Hash;
} sgt[MAXN * 4];
void Build(int now = 1, int left = 1, int right = n) {
  for (int i = left; i <= right; ++i) {
    sgt[now].num.push_back(app[i]);
  }
  sort(sgt[now].num.begin(), sgt[now].num.end());
  if (left == right) {
    return;
  }
  Build((now << 1), left, ((left + right) >> 1));
  Build((now << 1 | 1), ((left + right) >> 1) + 1, right);
}
int Query(int now_left, int now_right, int now = 1) {
  int left = lower_bound(sgt[now].num.begin(), sgt[now].num.end(), now_left) -
             sgt[now].num.begin();
  int right = upper_bound(sgt[now].num.begin(), sgt[now].num.end(), now_right) -
              sgt[now].num.begin() - 1;
  if (right < left) {
    return 0;
  }
  if (left == right) {
    return sgt[now].num[left];
  }
  int p = sgt[now].Hash[make_pair(left, right)];
  if (p) {
    return p;
  }
  int lson = Query(now_left, now_right, (now << 1)),
      rson = Query(now_left, now_right, (now << 1 | 1));
  if (!lson || !rson) {
    return sgt[now].Hash[make_pair(left, right)] = lson | rson;
  }
  use[++cnt] = Range(lson, rson);
  return sgt[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", &arr[i]);
    app[arr[i]] = i;
  }
  Build();
  int left, right;
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d", &left, &right);
    answer[i] = Query(left, right);
  }
  printf("%d\n", cnt);
  for (int i = n + 1; i <= cnt; ++i) {
    printf("%d %d\n", use[i].left, use[i].right);
  }
  for (int i = 1; i <= q; ++i) {
    printf("%d ", answer[i]);
  }
  return 0;
}