DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

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


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
inline int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline string get() {
  char ch[10];
  scanf("%s", ch);
  return ch;
}
template <class P, class Q>
inline P smin(P &a, Q b) {
  if (b < a) a = b;
  return a;
}
template <class P, class Q>
inline P smax(P &a, Q b) {
  if (a < b) a = b;
  return a;
}
const long long maxn = 1e5 + 10;
const long long base = 29;
const long long MAX_LG = 21;
const long long mod = 1e9 + 7;
const long long INF = 1e18;
const long long maxq = 2e5 + 10;
long long a[maxn], flat[4 * maxn], lazy[4 * maxn], stree[4 * maxn];
inline void build(long long p, long long l, long long r) {
  if (l == r) {
    flat[p] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(2 * p + 1, l, mid);
  build(2 * p + 2, mid + 1, r);
  flat[p] = 0;
}
inline void clear(long long p, long long l, long long r, long long val) {
  if (flat[p]) {
    stree[p] += abs(flat[p] - val) * (r - l + 1);
    lazy[p] += abs(flat[p] - val);
  } else {
    long long mid = (l + r) >> 1;
    clear(2 * p + 1, l, mid, val), clear(2 * p + 2, mid + 1, r, val);
    stree[p] = stree[2 * p + 1] + stree[2 * p + 2] + lazy[p] * (r - l + 1);
  }
}
inline void update(long long p, long long l, long long r, long long x,
                   long long y, long long val) {
  if (l > y || r < x) return;
  if (l >= x && r <= y) {
    clear(p, l, r, val);
    flat[p] = val;
    return;
  }
  long long mid = (l + r) >> 1;
  if (flat[p]) flat[2 * p + 1] = flat[2 * p + 2] = flat[p], flat[p] = 0;
  update(2 * p + 1, l, mid, x, y, val),
      update(2 * p + 2, mid + 1, r, x, y, val);
  stree[p] = stree[2 * p + 1] + stree[2 * p + 2] + lazy[p] * (r - l + 1);
}
inline long long get(long long p, long long l, long long r, long long x,
                     long long y) {
  if (l > y || r < x) return 0;
  if (l >= x && r <= y) return stree[p];
  long long mid = (l + r) >> 1;
  return get(2 * p + 1, l, mid, x, y) + get(2 * p + 2, mid + 1, r, x, y) +
         lazy[p] * (min(r, y) - max(l, x) + 1);
}
int32_t main() {
  long long n = in(), q = in();
  for (long long i = 0; i < n; i++) a[i] = i + 1;
  build(0, 0, n - 1);
  while (q--) {
    long long tp = in(), l = in() - 1, r = in() - 1;
    if (tp == 1) {
      long long x = in();
      update(0, 0, n - 1, l, r, x);
    } else
      cout << get(0, 0, n - 1, l, r) << "\n";
  }
}
