Recall that MEX of an array is a minimum non-negative integer that does not belong to the array. Examples:

  * for the array [0, 0, 1, 0, 2] MEX equals to 3 because numbers 0, 1 and 2 are presented in the array and 3 is the minimum non-negative integer not presented in the array; 
  * for the array [1, 2, 3, 4] MEX equals to 0 because 0 is the minimum non-negative integer not presented in the array; 
  * for the array [0, 1, 4, 3] MEX equals to 2 because 2 is the minimum non-negative integer not presented in the array. 



You are given an empty array a=[] (in other words, a zero-length array). You are also given a positive integer x.

You are also given q queries. The j-th query consists of one integer y_j and means that you have to append one element y_j to the array. The array length increases by 1 after a query.

In one move, you can choose any index i and set a_i := a_i + x or a_i := a_i - x (i.e. increase or decrease any element of the array by x). The only restriction is that a_i cannot become negative. Since initially the array is empty, you can perform moves only after the first query.

You have to maximize the MEX (minimum excluded) of the array if you can perform any number of such operations (you can even perform the operation multiple times with one element).

You have to find the answer after each of q queries (i.e. the j-th answer corresponds to the array of length j).

Operations are discarded before each query. I.e. the array a after the j-th query equals to [y_1, y_2, ..., y_j].

Input

The first line of the input contains two integers q, x (1 ≤ q, x ≤ 4 ⋅ 10^5) — the number of queries and the value of x.

The next q lines describe queries. The j-th query consists of one integer y_j (0 ≤ y_j ≤ 10^9) and means that you have to append one element y_j to the array.

Output

Print the answer to the initial problem after each query — for the query j print the maximum value of MEX after first j queries. Note that queries are dependent (the array changes after each query) but operations are independent between queries.

Examples

Input


7 3
0
1
2
2
0
0
10


Output


1
2
3
3
4
4
7


Input


4 3
1
2
1
2


Output


0
0
0
0

Note

In the first example:

  * After the first query, the array is a=[0]: you don't need to perform any operations, maximum possible MEX is 1. 
  * After the second query, the array is a=[0, 1]: you don't need to perform any operations, maximum possible MEX is 2. 
  * After the third query, the array is a=[0, 1, 2]: you don't need to perform any operations, maximum possible MEX is 3. 
  * After the fourth query, the array is a=[0, 1, 2, 2]: you don't need to perform any operations, maximum possible MEX is 3 (you can't make it greater with operations). 
  * After the fifth query, the array is a=[0, 1, 2, 2, 0]: you can perform a[4] := a[4] + 3 = 3. The array changes to be a=[0, 1, 2, 2, 3]. Now MEX is maximum possible and equals to 4. 
  * After the sixth query, the array is a=[0, 1, 2, 2, 0, 0]: you can perform a[4] := a[4] + 3 = 0 + 3 = 3. The array changes to be a=[0, 1, 2, 2, 3, 0]. Now MEX is maximum possible and equals to 4. 
  * After the seventh query, the array is a=[0, 1, 2, 2, 0, 0, 10]. You can perform the following operations: 
    * a[3] := a[3] + 3 = 2 + 3 = 5, 
    * a[4] := a[4] + 3 = 0 + 3 = 3, 
    * a[5] := a[5] + 3 = 0 + 3 = 3, 
    * a[5] := a[5] + 3 = 3 + 3 = 6, 
    * a[6] := a[6] - 3 = 10 - 3 = 7, 
    * a[6] := a[6] - 3 = 7 - 3 = 4. 
The resulting array will be a=[0, 1, 2, 5, 3, 6, 4]. Now MEX is maximum possible and equals to 7. 
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class tr1 {
	static PrintWriter out;
	static StringBuilder sb;
	static int inf = (int) 1e9;
	static long mod = (long) 998244353;
	static int[] si;
	static ArrayList<Integer> primes;
	static HashSet<Integer> pr;
	static int n, k, m;
	static int[] in;
	static HashMap<Integer, Integer> factors;
	static HashSet<Integer> f;
	static int[] fac, a, b;
	static int[] l, r;
	static int[][] memo;
	static int[] numc;
	static ArrayList<Integer>[] ad;
	static HashMap<Integer, Integer> hm;
	static pair[] ed;
	static TreeSet<Integer>[] np;
	static TreeMap<pair, Integer> tm;
	static int[] le, re;
	static char[] h;
	static pair[] mm;
	static boolean[] vis;
	static int y;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		int n=sc.nextInt();
		int min =-1;
		int x=sc.nextInt();
		int []mod=new int [x];
		for(int i=0;i<n;i++) {
			int a=sc.nextInt();
			mod[a%x]++;
			while(mod[(min+1)%x]>=((min+1)/x)+1)
				min++;
			///System.out.println(a+" "+Arrays.toString(mod)+" "+min);
			out.println(min+1);
		}
		out.flush();
	}

	static class pair implements Comparable<pair> {
		int node;
		int level;

		pair(int f, int t) {
			node = t;
			level = f;
		}

		public String toString() {
			return node + " " + level;
		}

		@Override
		public int compareTo(pair o) {
			return level - o.level;
		}
	}

	static class pair1 implements Comparable<pair1> {
		int x;
		int y;
		int z;

		pair1(int f, int t, int u) {
			x = f;
			y = t;
			z = u;
			// num = n;
		}

		public String toString() {
			return x + " " + y;
		}

		@Override
		public int compareTo(pair1 o) {
			if (x == o.x)
				return y - o.y;
			return x - o.x;
		}
	}

	static public class FenwickTree { // one-based DS

		int n;
		int[] ft;

		FenwickTree(int size) {
			n = size;
			ft = new int[n + 1];
			// for(int i=1;i<=n;i++)
			// ft[i]=i;
		}

		int rsq(int b) // O(log n)
		{
			int sum = 0;
			while (b > 0) {
				sum += ft[b];
				b -= b & -b;
			} // min?
			return sum;
		}

		int rsq(int a, int b) {
			return rsq(b) - rsq(a - 1);
		}

		void point_update(int k, int val) // O(log n), update = increment
		{
			while (k <= n) {
				// System.out.println(k+" "+val);
				ft[k] += val;
				k += k & -k;
			} // min?
		}

		int point_query(int idx) // c * O(log n), c < 1
		{
			int sum = ft[idx];
			if (idx > 0) {
				int z = idx ^ (idx & -idx);
				--idx;
				while (idx != z) {
					// System.out.println("oo");
					sum -= ft[idx];
					idx ^= idx & -idx;
				}
			}
			return sum;
		}

		void scale(int c) {
			for (int i = 1; i <= n; ++i)
				ft[i] *= c;
		}

		int findIndex(int cumFreq) {
			int msk = n;
			while ((msk & (msk - 1)) != 0)
				msk ^= msk & -msk; // msk will contain the MSB of n

			int idx = 0;
			while (msk != 0) {
				int tIdx = idx + msk;
				if (tIdx <= n && cumFreq >= ft[tIdx]) {
					idx = tIdx;
					cumFreq -= ft[tIdx];
				}
				msk >>= 1;
			}
			if (cumFreq != 0)
				return -1;
			return idx;
		}
	}

	static public class SegmentTree { // 1-based DS, OOP

		int N; // the number of elements in the array as a power of 2 (i.e. after padding)
		long[] array, sTree, lazy;

		SegmentTree(long[] in) {
			array = in;
			N = in.length - 1;
			sTree = new long[N << 1]; // no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new long[N << 1];
			build(1, 1, N);
		}

		void build(int node, int b, int e) // O(n)
		{
			if (b == e)
				sTree[node] = array[b];
			else {
				int mid = b + e >> 1;
				build(node << 1, b, mid);
				build(node << 1 | 1, mid + 1, e);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}
		}

		void update_point(int index, int val) // O(log n)
		{
			index += N - 1;
			sTree[index] += val;
			while (index > 1) {
				index >>= 1;
				sTree[index] = sTree[index << 1] + sTree[index << 1 | 1];
			}
		}

		void update_range(int i, int j, long val) // O(log n)
		{
			update_range(1, 1, N, i, j, val);
		}

		void update_range(int node, int b, int e, int i, int j, long val) {
			if (i > e || j < b)
				return;
			if (b >= i && e <= j) {
				sTree[node] += (e - b + 1) * val;
				lazy[node] += val;
			} else {
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node << 1, b, mid, i, j, val);
				update_range(node << 1 | 1, mid + 1, e, i, j, val);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}

		}

		void propagate(int node, int b, int mid, int e) {
			lazy[node << 1] += lazy[node];
			lazy[node << 1 | 1] += lazy[node];
			sTree[node << 1] += (mid - b + 1) * lazy[node];
			sTree[node << 1 | 1] += (e - mid) * lazy[node];
			lazy[node] = 0;
		}

		long query(int i, int j) {
			return query(1, 1, N, i, j);
		}

		long query(int node, int b, int e, int i, int j) // O(log n)
		{
			if (i > e || j < b)
				return 0;
			if (b >= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
			propagate(node, b, mid, e);
			long q1 = query(node << 1, b, mid, i, j);
			long q2 = query(node << 1 | 1, mid + 1, e, i, j);
			return q1 + q2;

		}

	}

	static long gcd(long a, long b) {

		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	public int[] merge(int[] d, int st, int e) {
		if (st > e)
			return null;
		if (e == st) {
			int[] ans = { d[e] };
			return ans;
		}
		int mid = (st + e) / 2;
		int[] ans = new int[e - st + 1];
		int[] ll = merge(d, st, mid);
		int[] rr = merge(d, mid + 1, e);
		if (ll == null)
			return rr;
		if (rr == null)
			return ll;
		int iver = 0;
		int idxl = 0;
		int idxr = 0;
		for (int i = st; i <= e; i++) {
			if (ll[idxl] < rr[idxr]) {

			}
		}
		return ans;
	}

	public static class pair2 implements Comparable<pair2> {
		int a;
		int idx;

		pair2(int a, int i) {
			this.a = a;
			idx = i;
		}

		public String toString() {
			return a + " " + idx;
		}

		@Override
		public int compareTo(pair2 o) {
			// TODO Auto-generated method stub
			return idx - o.idx;
		}
	}

	static long inver(long x) {
		int a = (int) x;
		long e = (mod - 2);
		long res = 1;
		while (e > 0) {
			if ((e & 1) == 1) {
				// System.out.println(res*a);
				res = (int) ((1l * res * a) % mod);
			}
			a = (int) ((1l * a * a) % mod);
			e >>= 1;
		}
		// out.println(res+" "+x);
		return res % mod;
	}

	static int atMostSum(int arr[], int n, int k) {
		int sum = 0;
		int cnt = 0, maxcnt = 0;

		for (int i = 0; i < n; i++) {

			// If adding current element doesn't
			// cross limit add it to current window
			if ((sum + arr[i]) <= k) {
				sum += arr[i];
				cnt++;
			}

			// Else, remove first element of current
			// window and add the current element
			else if (sum != 0) {
				sum = sum - arr[i - cnt] + arr[i];
			}

			// keep track of max length.
			maxcnt = Math.max(cnt, maxcnt);
		}
		return maxcnt;
	}

	public static int[] longestSubarray(int[] inp) {
		// array containing prefix sums up to a certain index i
		int[] p = new int[inp.length];
		p[0] = inp[0];
		for (int i = 1; i < inp.length; i++) {
			p[i] = p[i - 1] + inp[i];
		}

		// array Q from the description below
		int[] q = new int[inp.length];
		q[inp.length - 1] = p[inp.length - 1];
		for (int i = inp.length - 2; i >= 0; i--) {
			q[i] = Math.max(q[i + 1], p[i]);
		}
		int a = 0;
		int b = 0;
		int maxLen = 0;
		int curr;
		int[] res = new int[] { -1, -1 };
		while (b < inp.length) {
			curr = a > 0 ? q[b] - p[a - 1] : q[b];
			if (curr >= 0) {
				if (b - a > maxLen) {
					maxLen = b - a;
					res = new int[] { a, b };
				}
				b++;
			} else {
				a++;
			}
		}
		return res;
	}

	static void factor(int n) {
		if (si[n] == n) {
			f.add(n);
			return;
		}
		f.add(si[n]);
		factor(n / si[n]);
	}

	static void seive() {
		si = new int[100001];
		primes = new ArrayList<>();
		int N = 100001;
		si[1] = 1;
		for (int i = 2; i < N; i++) {
			if (si[i] == 0) {
				si[i] = i;
				primes.add(i);
			}
			for (int j = 0; j < primes.size() && primes.get(j) <= si[i] && (i * primes.get(j)) < N; j++)
				si[primes.get(j) * i] = primes.get(j);

		}
	}

	static class unionfind {
		int[] p;
		int[] size;
		int jum;

		unionfind(int n) {
			p = new int[n];
			size = new int[n];

			for (int i = 0; i < n; i++) {
				p[i] = i;
			}
			jum = n;
			Arrays.fill(size, 1);
		}

		int findSet(int v) {
			if (v == p[v])
				return v;
			return p[v] = findSet(p[v]);
		}

		boolean sameSet(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return true;
			return false;
		}

		int max() {
			int max = 0;
			for (int i = 0; i < size.length; i++)
				if (size[i] > max)
					max = size[i];
			return max;
		}

		void combine(int a, int b) {
			a = findSet(a);
			b = findSet(b);
			if (a == b)
				return;
			jum--;
			if (size[a] > size[b]) {
				p[b] = a;
				size[a] += size[b];

			} else {
				p[a] = b;
				size[b] += size[a];
			}
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}