#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll read() {
    ll x = 0, f = 1; char ch = getchar();
    while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
    while (isdigit(ch)) { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

#define ls x << 1
#define rs x << 1 | 1
const ll Max = 100033;
const ll mod = 1000000007;
ll n, k, m, tag[Max * 4], s[Max], ans;
ll a[Max], b[Max], g[Max * 2], L[Max], R[Max], f[Max];

void pushdown(ll x) {
    if (tag[x]) {
        tag[ls] = tag[rs] = tag[x];
        tag[x] = 0;
    }
}
void change(ll x, ll l, ll r, ll L, ll R, ll g) {
	if (L > R) return;
    if (L <= l && r <= R) {
        tag[x] = g;
        return;
    }
    pushdown(x);
    ll mid = (l + r) >> 1;
    if (L <= mid) change(ls, l, mid, L, R, g);
    if (R > mid) change(rs, mid + 1, r, L, R, g);
}
ll query(ll x, ll l, ll r, ll p) {
    if (l == r) return tag[x];
    pushdown(x);
    ll mid = (l + r) >> 1;
    if (p <= mid) return query(ls, l, mid, p);
    else return query(rs, mid + 1, r, p);
}
ll calc(ll x) {
    ll y = query(1, 1, m, x);
    return y ? f[y] + (g[L[y]] - g[x] + k) % k : y;
}

int main() {
    n = read(), k = read();
    for (ll i = 1; i <= n; i++) {
        a[i] = read(), b[i] = read();
        s[i] = s[i - 1] + a[i];
        if (b[i] == 1 && 2ll * a[i] > k) {
            puts("-1"); return 0;
        }
    }
    for (ll i = 1; i <= n; i++) {
        if (b[i] & 1) {
            R[i] = ((-2ll * s[i]) % k + k) % k;
            L[i] = ((-2ll * s[i - 1]) % k + k) % k;
        }
        else L[i] = 0, R[i] = k - 1;
        g[++m] = L[i], g[++m] = R[i];
    }
    sort(g + 1, g + m + 1);
    m = unique(g + 1, g + m + 1) - g - 1;
    for (ll i = n; i >= 1; i--) {
        L[i] = lower_bound(g + 1, g + m + 1, L[i]) - g;
        R[i] = lower_bound(g + 1, g + m + 1, R[i]) - g;
        f[i] = calc(L[i]);
        if (L[i] > R[i]) change(1, 1, m, R[i] + 1, L[i] - 1, i);
        else {
            change(1, 1, m, 1, L[i] - 1, i);
            change(1, 1, m, R[i] + 1, m, i);
        }
    }
    ans = f[1];
    for (ll i = 1; i <= m; i++)
        ans = min(ans, calc(i));
    cout << ans + s[n] * 2ll << endl;
    return 0;
}

### Prompt

Generate a CPP solution to the following problem:

There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll read() {
    ll x = 0, f = 1; char ch = getchar();
    while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
    while (isdigit(ch)) { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

#define ls x << 1
#define rs x << 1 | 1
const ll Max = 100033;
const ll mod = 1000000007;
ll n, k, m, tag[Max * 4], s[Max], ans;
ll a[Max], b[Max], g[Max * 2], L[Max], R[Max], f[Max];

void pushdown(ll x) {
    if (tag[x]) {
        tag[ls] = tag[rs] = tag[x];
        tag[x] = 0;
    }
}
void change(ll x, ll l, ll r, ll L, ll R, ll g) {
	if (L > R) return;
    if (L <= l && r <= R) {
        tag[x] = g;
        return;
    }
    pushdown(x);
    ll mid = (l + r) >> 1;
    if (L <= mid) change(ls, l, mid, L, R, g);
    if (R > mid) change(rs, mid + 1, r, L, R, g);
}
ll query(ll x, ll l, ll r, ll p) {
    if (l == r) return tag[x];
    pushdown(x);
    ll mid = (l + r) >> 1;
    if (p <= mid) return query(ls, l, mid, p);
    else return query(rs, mid + 1, r, p);
}
ll calc(ll x) {
    ll y = query(1, 1, m, x);
    return y ? f[y] + (g[L[y]] - g[x] + k) % k : y;
}

int main() {
    n = read(), k = read();
    for (ll i = 1; i <= n; i++) {
        a[i] = read(), b[i] = read();
        s[i] = s[i - 1] + a[i];
        if (b[i] == 1 && 2ll * a[i] > k) {
            puts("-1"); return 0;
        }
    }
    for (ll i = 1; i <= n; i++) {
        if (b[i] & 1) {
            R[i] = ((-2ll * s[i]) % k + k) % k;
            L[i] = ((-2ll * s[i - 1]) % k + k) % k;
        }
        else L[i] = 0, R[i] = k - 1;
        g[++m] = L[i], g[++m] = R[i];
    }
    sort(g + 1, g + m + 1);
    m = unique(g + 1, g + m + 1) - g - 1;
    for (ll i = n; i >= 1; i--) {
        L[i] = lower_bound(g + 1, g + m + 1, L[i]) - g;
        R[i] = lower_bound(g + 1, g + m + 1, R[i]) - g;
        f[i] = calc(L[i]);
        if (L[i] > R[i]) change(1, 1, m, R[i] + 1, L[i] - 1, i);
        else {
            change(1, 1, m, 1, L[i] - 1, i);
            change(1, 1, m, R[i] + 1, m, i);
        }
    }
    ans = f[1];
    for (ll i = 1; i <= m; i++)
        ans = min(ans, calc(i));
    cout << ans + s[n] * 2ll << endl;
    return 0;
}

```