Problem: The only difference between easy and hard versions is a number of elements in the array.

You are given an array a consisting of n integers. The value of the i-th element of the array is a_i.

You are also given a set of m segments. The j-th segment is [l_j; r_j], where 1 ≤ l_j ≤ r_j ≤ n.

You can choose some subset of the given set of segments and decrease values on each of the chosen segments by one (independently). For example, if the initial array a = [0, 0, 0, 0, 0] and the given segments are [1; 3] and [2; 4] then you can choose both of them and the array will become b = [-1, -2, -2, -1, 0].

You have to choose some subset of the given segments (each segment can be chosen at most once) in such a way that if you apply this subset of segments to the array a and obtain the array b then the value max_{i=1}^{n}b_i - min_{i=1}^{n}b_i will be maximum possible.

Note that you can choose the empty set.

If there are multiple answers, you can print any.

If you are Python programmer, consider using PyPy instead of Python when you submit your code.

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 300, 0 ≤ m ≤ 300) — the length of the array a and the number of segments, respectively.

The second line of the input contains n integers a_1, a_2, ..., a_n (-10^6 ≤ a_i ≤ 10^6), where a_i is the value of the i-th element of the array a.

The next m lines are contain two integers each. The j-th of them contains two integers l_j and r_j (1 ≤ l_j ≤ r_j ≤ n), where l_j and r_j are the ends of the j-th segment.

Output

In the first line of the output print one integer d — the maximum possible value max_{i=1}^{n}b_i - min_{i=1}^{n}b_i if b is the array obtained by applying some subset of the given segments to the array a.

In the second line of the output print one integer q (0 ≤ q ≤ m) — the number of segments you apply.

In the third line print q distinct integers c_1, c_2, ..., c_q in any order (1 ≤ c_k ≤ m) — indices of segments you apply to the array a in such a way that the value max_{i=1}^{n}b_i - min_{i=1}^{n}b_i of the obtained array b is maximum possible.

If there are multiple answers, you can print any.

Examples

Input


5 4
2 -2 3 1 2
1 3
4 5
2 5
1 3


Output


6
2
1 4 


Input


5 4
2 -2 3 1 4
3 5
3 4
2 4
2 5


Output


7
2
3 2 


Input


1 0
1000000


Output


0
0

Note

In the first example the obtained array b will be [0, -4, 1, 1, 2] so the answer is 6.

In the second example the obtained array b will be [2, -3, 1, -1, 4] so the answer is 7.

In the third example you cannot do anything so the answer is 0.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long INF = 1e12;
const int MOD = 1e9 + 7;
const unsigned long long BIT_FLAG_0 = (1 << 0);
const unsigned long long BIT_FLAG_1 = (1 << 1);
const unsigned long long BIT_FLAG_2 = (1 << 2);
const unsigned long long BIT_FLAG_3 = (1 << 3);
const unsigned long long BIT_FLAG_4 = (1 << 4);
const unsigned long long BIT_FLAG_5 = (1 << 5);
const unsigned long long BIT_FLAG_6 = (1 << 6);
const unsigned long long BIT_FLAG_7 = (1 << 7);
const long long dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const long long Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1},
                Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
void print() { cout << "\n"; }
template <class Head, class... Tail>
void print(Head &&head, Tail &&...tail) {
  cout << head;
  if (sizeof...(tail) != 0) cout << " ";
  print(forward<Tail>(tail)...);
}
template <class T>
void print(vector<T> &vec) {
  for (auto &a : vec) {
    cout << a;
    if (&a != &vec.back()) cout << " ";
  }
  cout << "\n";
}
template <class T>
void print(set<T> &set) {
  for (auto &a : set) {
    cout << a << " ";
  }
  cout << "\n";
}
template <class T>
void print(vector<vector<T>> &df) {
  for (auto &vec : df) {
    print(vec);
  }
}
long long power(long long a, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a;
    a *= a;
    n >>= 1;
  }
  return res;
}
long long comb(long long n, long long k) {
  vector<vector<long long>> v(n + 1, vector<long long>(n + 1, 0));
  for (long long i = 0; i < v.size(); i++) {
    v[i][0] = 1;
    v[i][i] = 1;
  }
  for (long long k = 1; k < v.size(); k++) {
    for (long long j = 1; j < k; j++) {
      v[k][j] = (v[k - 1][j - 1] + v[k - 1][j]);
    }
  }
  return v[n][k];
}
void add(long long &a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
__attribute__((constructor)) void faster_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
long long n, m;
signed main() {
  cin >> n >> m;
  vector<long long> A(n);
  for (long long ni = 0; ni < (long long)(n); ni++) {
    cin >> A[ni];
  }
  vector<pair<long long, long long>> edges;
  for (long long mi = 0; mi < (long long)(m); mi++) {
    long long l, r;
    cin >> l >> r;
    l--, r--;
    edges.push_back(make_pair(l, r));
  }
  vector<long long> ans;
  long long ansval = 0;
  for (long long ni = 0; ni < (long long)(n); ni++) {
    for (long long nj = 0; nj < (long long)(n); nj++) {
      long long mv = A[ni], Mv = A[nj];
      if (ni == nj) {
        continue;
      }
      vector<long long> tempans;
      long long cur = 1;
      for (auto ed : edges) {
        long long l = ed.first, r = ed.second;
        if ((l <= ni and ni <= r) and (l <= nj and nj <= r)) {
          cur++;
          continue;
        } else if ((l <= ni and ni <= r) and !(l <= nj and nj <= r)) {
          mv--;
          tempans.push_back(cur);
          cur++;
        } else {
          cur++;
          continue;
        }
      }
      if (Mv - mv > ansval) {
        ansval = Mv - mv;
        ans = tempans;
      }
    }
  }
  print(ansval);
  print((long long)((ans).size()));
  print(ans);
  return 0;
}