Problem: There are N non-negative integers written on the blackboard: A_1, ..., A_N.

Snuke can perform the following two operations at most K times in total in any order:

* Operation A: Replace each integer X on the blackboard with X divided by 2, rounded down to the nearest integer.
* Operation B: Replace each integer X on the blackboard with X minus 1. This operation cannot be performed if one or more 0s are written on the blackboard.



Find the number of the different possible combinations of integers written on the blackboard after Snuke performs the operations, modulo 1,000,000,007.

Constraints

* 1 \leq N \leq 200
* 1 \leq A_i \leq 10^{18}
* 1 \leq K \leq 10^{18}
* A_i and K are integers.

Input

Input is given from Standard Input in the following format:


N K
A_1 A_2 ... A_N


Output

Print the number of the different possible combinations of integers written on the blackboard after Snuke performs the operations, modulo 1,000,000,007.

Examples

Input

2 2
5 7


Output

6


Input

3 4
10 13 22


Output

20


Input

1 100
10


Output

11


Input

10 123456789012345678
228344079825412349 478465001534875275 398048921164061989 329102208281783917 779698519704384319 617456682030809556 561259383338846380 254083246422083141 458181156833851984 502254767369499613


Output

164286011

Solution (in CPP):

Solution:
#ifndef BZ
#ifndef __clang__
#pragma GCC optimize "-O3"
#endif
#endif
#include <bits/stdc++.h>

#define FASTIO
#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/

const ll MOD = 1e9 + 7;

ll n, k;
ll a[210];

int getmin(ll l, ll r) {
	if (l == r)
		return __builtin_popcountll(l);
	int ans = 0;
	for (int x = 60; x >= 0; --x) {
		if ((l >> x) != (r >> x)) {
			ans += 1;
			break;
		}
		else {
			ans += (l >> x) & 1;
		}
	}
	return min(ans, __builtin_popcountll(l));
}

map<vector<ll>, vector<pair<ll, ll> > > mm;
ll b[210];

int main() {
#ifdef FASTIO
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
	cin >> n >> k;
	for (int i = 0; i < n; ++i)
		cin >> a[i];
	sort(a, a + n);
	for (int x = 0; x <= 60; ++x) {
		vector<ll> vv;
		for (int i = 0; i < n; ++i)
			vv.push_back(a[i] % (1ll << x));
		vv.push_back((1ll << x) - 1);
		sort(ALL(vv));
		vv.resize(unique(ALL(vv)) - vv.begin());
		ll lst = 0;
		for (ll r: vv) {
			ll l = lst;
			int cnt = getmin(l, r);
			if (cnt + x > k || r > a[0])
				continue;
			for (int i = 0; i < n; ++i)
				b[i] = (a[i] - r) >> x;
			vector<ll> vv;
			for (int i = 1; i < n; ++i)
				vv.push_back(b[i] - b[0]);
			mm[vv].emplace_back(max(0ll, b[0] - (k - cnt - x)), b[0] + 1);
			lst = r + 1;
		}
	}
	ll ans = 0;
	for (auto x: mm) {
		auto &vv = x.second;
		sort(vv.begin(), vv.end());
		if (vv.empty())
			continue;
		ll cur = vv[0].first;
		for (auto p: vv) {
			if (p.first <= cur) {
				if (p.second > cur)
					ans += p.second - cur, cur = p.second;
			}
			else {
				ans += p.second - p.first;
				cur = p.second;
			}
		}
	}
	cout << ans % MOD << "\n";
	return 0;
}