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 PYTHON3):

Solution:
# after solving E1 it was quite easy to come up with the idea of lazy segtree

import sys

# a very nice implementation of a minimum segment tree with 
# some inspiration taken from https://codeforces.com/blog/entry/18051
# this implementation should be able to be modified to do pretty
# much anything one would want to do with segment trees apart from
# persistance.
# note that especially in python this implementation is much much better
# than most other approches because how slow python can be with function
# calls. /pajenegod

# currently it allows for two operations, both running in o(log n),
# 'add(l,r,value)' adds value to [l,r)
# 'find_min(l,r)' finds the index with the smallest value

big = 10**9

class super_seg:
    def __init__(self,data):
        n = len(data)
        m = 1
        while m<n: m *= 2
        
        self.n = n
        self.m = m
        self.data = [big]*(2*m)
        for i in range(n):
            self.data[i+m] = data[i]
        for i in reversed(range(m)):
            self.data[i] = min(self.data[2*i], self.data[2*i+1])
        self.query = [0]*(2*m)
    
    # push the query on seg_ind to its children
    def push(self,seg_ind):
        # let the children know of the queries
        q = self.query[seg_ind]

        self.query[2*seg_ind]   += q
        self.query[2*seg_ind+1] += q
        
        self.data[2*seg_ind]   += q
        self.data[2*seg_ind+1] += q

        # remove queries from seg_ind
        self.data[seg_ind] = min(self.data[2*seg_ind],self.data[2*seg_ind+1])
        self.query[seg_ind] = 0

    # updates the node seg_ind to know of all queries
    # applied to it via its ancestors
    def update(self,seg_ind):
        # find all indecies to be updated
        seg_ind //= 2
        inds = []
        while seg_ind>0:
            inds.append(seg_ind)
            seg_ind//=2
       
        # push the queries down the segment tree
        for ind in reversed(inds):
            self.push(ind)

    # make the changes to seg_ind be known to its ancestors
    def build(self,seg_ind):
        seg_ind//=2
        while seg_ind>0:
            self.data[seg_ind] = min(self.data[2*seg_ind], self.data[2*seg_ind+1]) + self.query[seg_ind]
            seg_ind //= 2

    # lazily add value to [l,r)
    def add(self,l,r,value):
        l += self.m
        r += self.m
        
        l0 = l
        r0 = r

        while l<r:
            if l%2==1:
                self.query[l]+= value
                self.data[l] += value
                l+=1
            if r%2==1:
                r-=1
                self.query[r]+= value
                self.data[r] += value
            l//=2
            r//=2

        # tell all nodes above of the updated
        # area of the updates
        self.build(l0)
        self.build(r0-1)
    
    # min of data[l,r)
    def min(self,l,r):
        l += self.m
        r += self.m

        # apply all the lazily stored queries
        self.update(l)
        self.update(r-1)
        
        segs = []
        while l<r:
            if l%2==1:
                segs.append(l)
                l+=1
            if r%2==1:
                r-=1
                segs.append(r)
            l//=2
            r//=2

        return min(self.data[ind] for ind in segs)


    # find index of smallest value in data[l,r)
    def find_min(self,l,r):
        l += self.m
        r += self.m
        
        # apply all the lazily stored queries
        self.update(l)
        self.update(r-1)

        segs = []
        while l<r:
            if l%2==1:
                segs.append(l)
                l+=1
            if r%2==1:
                r-=1
                segs.append(r)
            l//=2
            r//=2

        ind = min(segs, key=lambda i:self.data[i])
        mini = self.data[ind]
       
        # dig down in search of mini
        while ind<self.m:
            self.push(ind)

            if self.data[2*ind]==mini:
                ind *= 2
            else:
                ind = 2*ind+1
        
        return ind-self.m,mini


n,m = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]

inter = []
update = [[] for _ in range(n+1)]
for _ in range(m):
    l,r = [int(x) for x in input().split()]
    l -= 1
    inter.append((l,r))
    update[l].append((l,r))
    update[r].append((l,r))

Aneg = super_seg([-a for a in A])
besta = -1
besta_ind = -1

active_intervals = 0
for i in range(n):
    for l,r in update[i]:
        Aneg.add(l,r,1 if l==i else -1)
        active_intervals += 1 if l==i else -1 
    Amax = -Aneg.data[1]
    Ai = A[i] - active_intervals
    if Amax-Ai>besta:
        besta = Amax-Ai
        besta_ind = i

ints = [i for i in range(m) if inter[i][0]<=besta_ind<inter[i][1]]
print(besta)
print(len(ints))
print(*[x+1 for x in ints])